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2 As x → 2−, f (x) → ∞ As x → 2+, f (x) → −∞ Hole at −3, 7 5 Horizontal asymptote: y = 1 As x → −∞, f (x) → 1+ As x → ∞, f (x) → 1− y 1 −7−6−5−4−3−2−1 y 5 4 3 2 1 −5 −4 −3 −2 −1 1 2 3 4 5 x −1 −2 −3 −4 −5 5 4 3 2 1 y −5 −4 −3 −2 −1 1 2 3 4 5 x −1 −2 −3 −4 −5 338 Rational Functions 10. f (x) = = 3x2 − 5x − 2 x2 − 9 (3x + 1)(x − 2) (x + 3)(x − 3) Domain: (−∞, −3) ∪ (−3, 3) ∪ (3, ∞) x-intercepts: − 1 3, 0, (2, 0) y-intercept: 0, 2 9 Vertical asymptotes: x = −3, x = 3 As x → −3−, f (x) → ∞ As x → −3+, f (x) → −∞ As x → 3−, f (x) → −∞ As x → 3+, f (x) → ∞ No holes in the graph Horizontal asymptote: y = 3 As x → −∞, f (x) → 3+ As x → ∞, f (x) → 3− 11. f (x) = x2 − x − 6 x + 1 = (x − 3)(x + 2) x + 1 Domain: (−∞, −1) ∪ (−1, ∞) x-intercepts: (−2, 0), (3, 0) y-intercept: (0, −6) Vertical asymptote: x = −1 As x → −1−, f (x) → ∞ As x → −1+, f (x) → −∞ Slant asymptote: y = x − 2 As x → −∞, the graph is above y = x − 2 As x → ∞, the graph is below y = x − 2 12. f (
x) = x2 − x 3 − x = x(x − 1) 3 − x Domain: (−∞, 3) ∪ (3, ∞) x-intercepts: (0, 0), (1, 0) y-intercept: (0, 0) Vertical asymptote: x = 3 As x → 3−, f (x) → ∞ As x → 3+, f (x) → −∞ Slant asymptote: y = −x − 2 As x → −∞, the graph is above y = −x − 2 As x → ∞, the graph is below y = −9−8−7−7−6−5−4−3−2−1 −1 −3 −4 −5 −6 −7 −8 −4−3−2−1 1 2 3 4 x −2 −4 −6 y 8 6 4 2 −8−7−6−5−4−3−2−1 − 10 x −4 −6 −8 −10 −12 −14 −16 −18 4.2 Graphs of Rational Functions 339 13. f (x) = x3 + 2x2 + x x2 − x − 2 = x(x + 1) x − 2 x = −1 Domain: (−∞, −1) ∪ (−1, 2) ∪ (2, ∞) x-intercept: (0, 0) y-intercept: (0, 0) Vertical asymptote: x = 2 As x → 2−, f (x) → −∞ As x → 2+, f (x) → ∞ Hole at (−1, 0) Slant asymptote: y = x + 3 As x → −∞, the graph is below y = x + 3 As x → ∞, the graph is above y = x + 3 14. f (x) = −x3 + 4x x2 − 9 Domain: (−∞, −3) ∪ (−3, 3) ∪ (3, ∞) x-intercepts: (−2, 0), (0, 0), (2, 0) y-intercept: (0, 0) Vertical asymptotes: x = −3, x = 3 As x → −3−, f (x) → ∞ As x → −3+,
f (x) → −∞ As x → 3−, f (x) → ∞ As x → 3+, f (x) → −∞ Slant asymptote: y = −x As x → −∞, the graph is above y = −x As x → ∞, the graph is below y = −x 15. f (x) = x3 − 2x2 + 3x 2x2 + 2 Domain: (−∞, ∞) x-intercept: (0, 0) y-intercept: (0, 0) Slant asymptote: y = 1 As x → −∞, the graph is below y = 1 As x → ∞, the graph is above 18 16 14 12 10 8 6 4 2 −9−8−7−6−5−4−3−2−1 −4 −6 −8 −10 y 7 6 5 4 3 2 1 −6−5−4−3−2−1 −1 1 2 3 4 5 6 x −2 −3 −4 −5 −6 −7 y 2 1 −4 −3 −2 −1 1 2 3 4 x −1 −2 340 16. f (x) = x2 − 2x + 1 x3 + x2 − 2x Domain: (−∞, −2) ∪ (−2, 0) ∪ (0, 1) ∪ (1, ∞) f (x(x + 2) No x-intercepts No y-intercepts Vertical asymptotes: x = −2 and x = 0 As x → −2−, f (x) → −∞ As x → −2+, f (x) → ∞ As x → 0−, f (x) → ∞ As x → 0+, f (x) → −∞ Hole in the graph at (1, 0) Horizontal asymptote: y = 0 As x → −∞, f (x) → 0− As x → ∞, f (x) → 0+ 17. f (x) = 1 x − 2 Shift the graph of y = to the right 2 units. 1 x 18. g(x) = 1 − 3 x Rational Functions y 5 4 3 2 1 −5 −4 −3 −2 −1 1 2 3 4 5 x −1 −2 −3 −4
−5 y 3 2 1 −1 1 2 3 4 5 x −1 −2 − Vertically stretch the graph of y = by a factor of 3. 3 x Reﬂect the graph of y = about the x-axis. Shift the graph of y = − up 1 unit. 3 x −6−5−4−3−2−1 −1 1 2 3 4 5 6 x −2 −3 −4 −5 4.2 Graphs of Rational Functions 341 19. h(x) = −2x + 1 x Shift the graph of y = down 2 units. = −2 + 1 x 1 x y 1 −3 −2 −1 1 2 3 x −1 −2 −3 −4 −3 −2 −1 1 2 3 4 5 x −1 20. j(x) = = 3 − 3x − 7 x − 2 Shift the graph of y = to the right 2 units. 1 x − 2 1 x Reﬂect the graph of y = about the x-axis. Shift the graph of y = − up 3 units. 1 x − 2 1 x − 2 342 Rational Functions 4.3 Rational Inequalities and Applications In this section, we solve equations and inequalities involving rational functions and explore associated application problems. Our ﬁrst example showcases the critical diﬀerence in procedure between solving a rational equation and a rational inequality. Example 4.3.1. 1. Solve x3 − 2x +. 2. Solve x3 − 2x +. 3. Use your calculator to graphically check your answers to 1 and 2. Solution. 1. To solve the equation, we clear denominators x3 − 2x + 1 x − 1 = x3 − 2x + 1 x − 1 · 2(x − 1(x − 1) 2x3 − 4x + 2 = x2 − 3x + 2 2x3 − x2 − x = 0 x(2x + 1)(x − 1, 1 expand factor Since we cleared denominators, we need to check for extraneous solutions. Sure enough, we see that x = 1 does not satisfy the original equation and must be discarded. Our solutions are x = − 1 2 and x = 0. 2. To solve the inequality, it may be tempting to begin as we did with the equation − namely by multiplying both sides by the quantity (x − 1). The problem is that
, depending on x, (x − 1) may be positive (which doesn’t aﬀect the inequality) or (x − 1) could be negative (which would reverse the inequality). Instead of working by cases, we collect all of the terms on one side of the inequality with 0 on the other and make a sign diagram using the technique given on page 321 in Section 4.2. x3 − 2x + 1 x − 1 1 2 − x3 − 2x + x3 − 2x + 1 − x(x − 1) + 1(2(x − 1)) 2(x − 1) ≥ 0 ≥ 0 get a common denominator expand 2x3 − x2 − x 2x − 2 4.3 Rational Inequalities and Applications 343 Viewing the left hand side as a rational function r(x) we make a sign diagram. The only value excluded from the domain of r is x = 1 which is the solution to 2x − 2 = 0. The zeros of r are the solutions to 2x3 − x2 − x = 0, which we have already found to be x = 0, x = − 1 2 and x = 1, the latter was discounted as a zero because it is not in the domain. Choosing test values in each test interval, we construct the sign diagram below. (+) 0 (−) 0 (+) (+) − 1 2 0 1 We are interested in where r(x) ≥ 0. We ﬁnd r(x) > 0, or (+), on the intervals −∞, − 1 2 (0, 1) and (1, ∞). We add to these intervals the zeros of r, − 1 −∞, − 1 2, 2 and 0, to get our ﬁnal solution: ∪ [0, 1) ∪ (1, ∞). 3. Geometrically, if we set f (x) = x3−2x+1 and g(x) = 1 2 x − 1, the solutions to f (x) = g(x) are the x-coordinates of the points where the graphs of y = f (x) and y = g(x) intersect. The solution to f (x) ≥ g(x) represents not only where the graphs meet, but the intervals over which the graph of y = f (x) is above (>) the graph of g
(x). We obtain the graphs below. x−1 The ‘Intersect’ command conﬁrms that the graphs cross when x = − 1 2 and x = 0. It is clear from the calculator that the graph of y = f (x) is above the graph of y = g(x) on −∞, − 1 2 as well as on (0, ∞). According to the calculator, our solution is then −∞, − 1 ∪ [0, ∞) 2 which almost matches the answer we found analytically. We have to remember that f is not deﬁned at x = 1, and, even though it isn’t shown on the calculator, there is a hole1 in the graph of y = f (x) when x = 1 which is why x = 1 is not part of our ﬁnal answer. Next, we explore how rational equations can be used to solve some classic problems involving rates. Example 4.3.2. Carl decides to explore the Meander River, the location of several recent Sasquatch sightings. From camp, he canoes downstream ﬁve miles to check out a purported Sasquatch nest. Finding nothing, he immediately turns around, retraces his route (this time traveling upstream), 1There is no asymptote at x = 1 since the graph is well behaved near x = 1. According to Theorem 4.1, there must be a hole there. 344 Rational Functions and returns to camp 3 hours after he left. If Carl canoes at a rate of 6 miles per hour in still water, how fast was the Meander River ﬂowing on that day? Solution. We are given information about distances, rates (speeds) and times. The basic principle relating these quantities is: distance = rate · time The ﬁrst observation to make, however, is that the distance, rate and time given to us aren’t ‘compatible’: the distance given is the distance for only part of the trip, the rate given is the speed Carl can canoe in still water, not in a ﬂowing river, and the time given is the duration of the entire trip. Ultimately, we are after the speed of the river, so let’s call that R measured in miles per hour to be consistent with the other rate given to us. To get started, let’s divide
the trip into its two parts: the initial trip downstream and the return trip upstream. For the downstream trip, all we know is that the distance traveled is 5 miles. distance downstream = rate traveling downstream · time traveling downstream 5 miles = rate traveling downstream · time traveling downstream Since the return trip upstream followed the same route as the trip downstream, we know that the distance traveled upstream is also 5 miles. distance upstream = rate traveling upstream · time traveling upstream 5 miles = rate traveling upstream · time traveling upstream We are told Carl can canoe at a rate of 6 miles per hour in still water. How does this ﬁgure into the rates traveling upstream and downstream? The speed the canoe travels in the river is a combination of the speed at which Carl can propel the canoe in still water, 6 miles per hour, and the speed of the river, which we’re calling R. When traveling downstream, the river is helping Carl along, so we add these two speeds: rate traveling downstream = rate Carl propels the canoe + speed of the river hour + R miles = 6 miles hour So our downstream speed is (6 + R) miles for the downstream part of the trip, we get: hour. Substituting this into our ‘distance-rate-time’ equation 5 miles = rate traveling downstream · time traveling downstream 5 miles = (6 + R) miles hour · time traveling downstream When traveling upstream, Carl works against the current. Since the canoe manages to travel upstream, the speed Carl can canoe in still water is greater than the river’s speed, so we subtract the river’s speed from Carl’s canoing speed to get: rate traveling upstream = rate Carl propels the canoe − river speed = 6 miles hour − R miles hour Proceeding as before, we get 4.3 Rational Inequalities and Applications 345 5 miles = rate traveling upstream · time traveling upstream 5 miles = (6 − R) miles hour · time traveling upstream The last piece of information given to us is that the total trip lasted 3 hours. If we let tdown denote the time of the downstream trip and tup the time of the upstream trip, we have: tdown+tup = 3 hours. Substituting tdown and tup into the ‘distance-rate-time’ equations, we get (suppressing the units) three equations in three unknowns:2    E1 (6 + R) tdown = 5 (
6 − R) tup = 5 E2 tdown + tup = 3 E3 Since we are ultimately after R, we need to use these three equations to get at least one equation involving only R. To that end, we solve E1 for tdown by dividing both sides3 by the quantity (6+R) to get tdown = 5 6−R. Substituting these into E3, we get:4 6+R. Similarly, we solve E2 for tup and get tup =. Clearing denominators, we get 5(6 − R) + 5(6 + R) = 3(6 + R)(6 − R) which reduces to R2 = 16. We ﬁnd R = ±4, and since R represents the speed of the river, we choose R = 4. On the day in question, the Meander River is ﬂowing at a rate of 4 miles per hour. One of the important lessons to learn from Example 4.3.2 is that speeds, and more generally, rates, are additive. As we see in our next example, the concept of rate and its associated principles can be applied to a wide variety of problems - not just ‘distance-rate-time’ scenarios. Example 4.3.3. Working alone, Taylor can weed the garden in 4 hours. If Carl helps, they can weed the garden in 3 hours. How long would it take for Carl to weed the garden on his own? Solution. The key relationship between work and time which we use in this problem is: amount of work done = rate of work · time spent working We are told that, working alone, Taylor can weed the garden in 4 hours. In Taylor’s case then: amount of work Taylor does = rate of Taylor working · time Taylor spent working 1 garden = (rate of Taylor working) · (4 hours) So we have that the rate Taylor works is 1 garden together, Taylor and Carl can weed the garden in just 3 hours. We have: 4 hours = 1 garden hour. We are also told that when working 4 2This is called a system of equations. No doubt, you’ve had experience with these things before, and we will study systems in greater detail in Chapter 8. 3While we usually discourage dividing both sides of an equation by a variable expression, we know (6 + R) = 0 since otherwise we couldn’t possibly multiply it by tdown and get
5. 4The reader is encouraged to verify that the units in this equation are the same on both sides. To get you started, the units on the ‘3’ is ‘hours.’ 346 Rational Functions amount of work done together = rate of working together · time spent working together 1 garden = (rate of working together) · (3 hours) garden From this, we ﬁnd that the rate of Taylor and Carl working together is 1 garden hour. We are asked to ﬁnd out how long it would take for Carl to weed the garden on his own. Let us call this unknown t, measured in hours to be consistent with the other times given to us in the problem. Then: 3 hours = 1 3 amount of work Carl does = rate of Carl working · time Carl spent working 1 garden = (rate of Carl working) · (t hours) In order to ﬁnd t, we need to ﬁnd the rate of Carl working, so let’s call this quantity R, with units garden hour. Using the fact that rates are additive, we have: rate working together = rate of Taylor working + rate of Carl working hour + R garden hour = 1 garden garden hour 1 3 4 so that R = 1 12 garden hour. Substituting this into our ‘work-rate-time’ equation for Carl, we get: 1 garden = (rate of Carl working) · (t hours) 1 12 garden hour · (t hours) 1 garden = Solving 1 = 1 12 t, we get t = 12, so it takes Carl 12 hours to weed the garden on his own.5 As is common with ‘word problems’ like Examples 4.3.2 and 4.3.3, there is no short-cut to the answer. We encourage the reader to carefully think through and apply the basic principles of rate to each (potentially diﬀerent!) situation. It is time well spent. We also encourage the tracking of units, especially in the early stages of the problem. Not only does this promote uniformity in the units, it also serves as a quick means to check if an equation makes sense.6 Our next example deals with the average cost function, ﬁrst introduced on page 82, as applied to PortaBoy Game systems from Example 2.1.5 in Section 2.1. Example 4.3.4. Given a cost function C(x
), which returns the total cost of producing x items, recall that the average cost function, C(x) = C(x) computes the cost per item when x items are x produced. Suppose the cost C, in dollars, to produce x PortaBoy game systems for a local retailer is C(x) = 80x + 150, x ≥ 0. 1. Find an expression for the average cost function C(x). 2. Solve C(x) < 100 and interpret. 5Carl would much rather spend his time writing open-source Mathematics texts than gardening anyway. 6In other words, make sure you don’t try to add apples to oranges! 4.3 Rational Inequalities and Applications 347 3. Determine the behavior of C(x) as x → ∞ and interpret. Solution. 1. From C(x) = C(x) x, we obtain C(x) = 80x+150 x. The domain of C is x ≥ 0, but since x = 0 causes problems for C(x), we get our domain to be x > 0, or (0, ∞). 2. Solving C(x) < 100 means we solve 80x+150 x < 100. We proceed as in the previous example. 80x + 150 x < 100 80x + 150 x − 100 < 0 80x + 150 − 100x x 150 − 20x x < 0 common denominator < 0 If we take the left hand side to be a rational function r(x), we need to keep in mind that the applied domain of the problem is x > 0. This means we consider only the positive half of the number line for our sign diagram. On (0, ∞), r is deﬁned everywhere so we need only look for zeros of r. Setting r(x) = 0 gives 150 − 20x = 0, so that x = 15 2 = 7.5. The test intervals on our domain are (0, 7.5) and (7.5, ∞). We ﬁnd r(x) < 0 on (7.5, ∞). 0 (+) 0 (−) 7.5 In the context of the problem, x represents the number of PortaBoy games systems produced and C(x) is the average cost to produce each system. Solving C(x) < 100 means we are trying to ﬁnd how many systems we need to produce
so that the average cost is less than $100 per system. Our solution, (7.5, ∞) tells us that we need to produce more than 7.5 systems to achieve this. Since it doesn’t make sense to produce half a system, our ﬁnal answer is [8, ∞). 3. When we apply Theorem 4.2 to C(x) we ﬁnd that y = 80 is a horizontal asymptote to the graph of y = C(x). To more precisely determine the behavior of C(x) as x → ∞, we ﬁrst use long division7 and rewrite C(x) = 80 + 150 x → 0+, which means C(x) ≈ 80 + very small (+). Thus the average cost per system is getting closer to $80 per system. x = 0, which is impossible, so we conclude that C(x) > 80 for all x > 0. This means that the average cost per system is always greater than $80 per system, but the average cost is approaching this amount as more and more systems are produced. Looking back at Example 2.1.5, we realize $80 is the variable cost per system − If we set C(x) = 80, we get 150 x. As x → ∞, 150 7In this case, long division amounts to term-by-term division. 348 Rational Functions the cost per system above and beyond the ﬁxed initial cost of $150. Another way to interpret our answer is that ‘inﬁnitely’ many systems would need to be produced to eﬀectively ‘zero out’ the ﬁxed cost. Our next example is another classic ‘box with no top’ problem. Example 4.3.5. A box with a square base and no top is to be constructed so that it has a volume of 1000 cubic centimeters. Let x denote the width of the box, in centimeters as seen below. height depth width, x 1. Express the height h in centimeters as a function of the width x and state the applied domain. 2. Solve h(x) ≥ x and interpret. 3. Find and interpret the behavior of h(x) as x → 0+ and as x → ∞. 4. Express the surface area S of the box as a function of x and state the applied domain. 5
. Use a calculator to approximate (to two decimal places) the dimensions of the box which minimize the surface area. Solution. 1. We are told that the volume of the box is 1000 cubic centimeters and that x represents the width, in centimeters. From geometry, we know Volume = width × height × depth. Since the base of the box is a square, the width and the depth are both x centimeters. Using h for the height, we have 1000 = x2h, so that h = 1000 x2 As for the applied domain, in order for there to be a box at all, x > 0, and since every such choice of x will return a positive number for the height h we have no other restrictions and conclude our domain is (0, ∞). x2. Using function notation,8 h(x) = 1000 2. To solve h(x) ≥ x, we proceed as before and collect all nonzero terms on one side of the inequality in order to use a sign diagram. 8That is, h(x) means ‘h of x’, not ‘h times x’ here. 4.3 Rational Inequalities and Applications 349 h(x) ≥ x 1000 x2 ≥ x 1000 x2 − x ≥ 0 1000 − x3 x2 ≥ 0 common denominator We consider the left hand side of the inequality as our rational function r(x). We see r is undeﬁned at x = 0, but, as in the previous example, the applied domain of the problem is x > 0, so we are considering only the behavior of r on (0, ∞). The sole zero of r comes when 1000 − x3 = 0, which is x = 10. Choosing test values in the intervals (0, 10) and (10, ∞) gives the following diagram. (+) 0 (−) 0 10 We see r(x) > 0 on (0, 10), and since r(x) = 0 at x = 10, our solution is (0, 10]. In the context of the problem, h represents the height of the box while x represents the width (and depth) of the box. Solving h(x) ≥ x is tantamount to ﬁnding the values of x which result in a box where the height is at least as big as the width (and, in this case, depth.) Our answer tells us the width of the box can be at most 10
centimeters for this to happen. 3. As x → 0+, h(x) = 1000 x2 → ∞. This means that the smaller the width x (and, in this case, depth), the larger the height h has to be in order to maintain a volume of 1000 cubic centimeters. As x → ∞, we ﬁnd h(x) → 0+, which means that in order to maintain a volume of 1000 cubic centimeters, the width and depth must get bigger as the height becomes smaller. 4. Since the box has no top, the surface area can be found by adding the area of each of the sides to the area of the base. The base is a square of dimensions x by x, and each side has dimensions x by h. We get the surface area, S = x2 + 4xh. To get S as a function of x, we. Hence, as a function of x, S(x) = x2 + 4000 substitute h = 1000 x. x2 The domain of S is the same as h, namely (0, ∞), for the same reasons as above. to obtain S = x2 + 4x 1000 x2 5. A ﬁrst attempt at the graph of y = S(x) on the calculator may lead to frustration. Chances are good that the ﬁrst window chosen to view the graph will suggest y = S(x) has the x-axis as a horizontal asymptote. From the formula S(x) = x2 + 4000 x, however, we get S(x) ≈ x2 as x → ∞, so S(x) → ∞. Readjusting the window, we ﬁnd S does possess a relative minimum at x ≈ 12.60. As far as we can tell,9 this is the only relative extremum, so it is the absolute minimum as well. This means that the width and depth of the box should each measure 9without Calculus, that is... 350 Rational Functions approximately 12.60 centimeters. To determine the height, we ﬁnd h(12.60) ≈ 6.30, so the height of the box should be approximately 6.30 centimeters. 4.3.1 Variation In many instances in the sciences, rational functions are encountered as a result of fundamental natural laws which are typically a result of assuming certain basic relationships between variables. These basic relationships are summarized in the
deﬁnition below. Deﬁnition 4.5. Suppose x, y and z are variable quantities. We say y varies directly with (or is directly proportional to) x if there is a constant k such that y = kx. y varies inversely with (or is inversely proportional to) x if there is a constant k such that y = k x. z varies jointly with (or is jointly proportional to) x and y if there is a constant k such that z = kxy. The constant k in the above deﬁnitions is called the constant of proportionality. Example 4.3.6. Translate the following into mathematical equations using Deﬁnition 4.5. 1. Hooke’s Law: The force F exerted on a spring is directly proportional the extension x of the spring. 2. Boyle’s Law: At a constant temperature, the pressure P of an ideal gas is inversely propor- tional to its volume V. 3. The volume V of a right circular cone varies jointly with the height h of the cone and the square of the radius r of the base. 4. Ohm’s Law: The current I through a conductor between two points is directly proportional to the voltage V between the two points and inversely proportional to the resistance R between the two points. 4.3 Rational Inequalities and Applications 351 5. Newton’s Law of Universal Gravitation: Suppose two objects, one of mass m and one of mass M, are positioned so that the distance between their centers of mass is r. The gravitational force F exerted on the two objects varies directly with the product of the two masses and inversely with the square of the distance between their centers of mass. Solution. 1. Applying the deﬁnition of direct variation, we get F = kx for some constant k. 2. Since P and V are inversely proportional, we write P = k V. 3. There is a bit of ambiguity here. It’s clear that the volume and the height of the cone are represented by the quantities V and h, respectively, but does r represent the radius of the base or the square of the radius of the base? It is the former. Usually, if an algebraic operation is speciﬁed (like squaring), it is meant to be expressed in the formula. We apply Deﬁnition 4
.5 to get V = khr2. 4. Even though the problem doesn’t use the phrase ‘varies jointly’, it is implied by the fact that the current I is related to two diﬀerent quantities. Since I varies directly with V but inversely with R, we write I = kV R. 5. We write the product of the masses mM and the square of the distance as r2. We have that F varies directly with mM and inversely with r2, so F = kmM r2. In many of the formulas in the previous example, more than two varying quantities are related. In practice, however, usually all but two quantities are held constant in an experiment and the data collected is used to relate just two of the variables. Comparing just two varying quantities allows us to view the relationship between them as functional, as the next example illustrates. Example 4.3.7. According to this website the actual data relating the volume V of a gas and its pressure P used by Boyle and his assistant in 1662 to verify the gas law that bears his name is given below. V P V P 48 46 44 42 40 38 36 34 32 30 28 26 24 29.13 30.56 31.94 33.5 35.31 37 39.31 41.63 44.19 47.06 50.31 54.31 58.81 23 61.31 22 21 20 19 18 17 16 15 14 13 12 64.06 67.06 70.69 74.13 77.88 82.75 87.88 93.06 100.44 107.81 117.56 1. Use your calculator to generate a scatter diagram for these data using V as the independent variable and P as the dependent variable. Does it appear from the graph that P is inversely proportional to V? Explain. 2. Assuming that P and V do vary inversely, use the data to approximate the constant of proportionality. 352 Rational Functions 3. Use your calculator to determine a ‘Power Regression’ for this data10 and use it verify your results in 1 and 2. Solution. 1. If P really does vary inversely with V, then P = k V for some constant k. From the data plot, the points do seem to lie along a curve like y = k x. 2. To determine the constant of proportionality, we note that from P = k V, we get k = P V. Multiplying each of the
volume numbers times each of the pressure numbers,11 we produce a number which is always approximately 1400. We suspect that P = 1400 V. Graphing y = 1400 along with the data gives us good reason to believe our hypotheses that P and V are, in fact, inversely related. x The graph of the data The data with y = 1400 x 3. After performing a ‘Power Regression’, the calculator ﬁts the data to the curve y = axb where a ≈ 1400 and b ≈ −1 with a correlation coeﬃcient which is darned near perfect.12 In other words, y = 1400x−1 or y = 1400 x, as we guessed. 10We will talk more about this in the coming chapters. 11You can use tell the calculator to do this arithmetic on the lists and save yourself some time. 12We will revisit this example once we have developed logarithms in Chapter 6 to see how we can actually ‘linearize’ this data and do a linear regression to obtain the same result. 4.3 Rational Inequalities and Applications 353 4.3.2 Exercises In Exercises 1 - 6, solve the rational equation. Be sure to check for extraneous solutions. 1. 3. 5. x 5x + 4 = 3 = x2 − 3 x2 − x2 − 2x + 1 x3 + x2 − 2x = 1 2. 4. 6. 3x − 1 x2 + 1 = 1 2x + 17 x + 1 = x + 5 −x3 + 4x x2 − 9 = 4x In Exercises 7 - 20, solve the rational inequality. Express your answer using interval notation. 7. 1 x + 2 ≥ 0 4x x2 + 4 ≥ 0 ≥ 0 x3 + 2x2 + x x2 − x − 2 2x + 17 x + 1 > x + 5 10. 13. 16. 19. 8. 11. 14. 17. ≤ 0 x − 3 x + 2 x2 − x − 12 x2 + x − 6 x2 + 5x + 6 x2 − 1 −x3 + 4x x2 − 9 > 0 > 0 ≥ 4x 9. x x2 − 1 > 0 12. 15. 18. 3x2 − 5x − 2 x2 − 9 < 0 3x − 1 x2 + 1 1 x2 + 1 ≤ 1
< 0 x4 − 4x3 + x2 − 2x − 15 x3 − 4x2 ≥ x 20. 5x3 − 12x2 + 9x + 10 x2 − 1 ≥ 3x − 1 21. Carl and Mike start a 3 mile race at the same time. If Mike ran the race at 6 miles per hour and ﬁnishes the race 10 minutes before Carl, how fast does Carl run? 22. One day, Donnie observes that the wind is blowing at 6 miles per hour. A unladen swallow nesting near Donnie’s house ﬂies three quarters of a mile down the road (in the direction of the wind), turns around, and returns exactly 4 minutes later. What is the airspeed of the unladen swallow? (Here, ‘airspeed’ is the speed that the swallow can ﬂy in still air.) 23. In order to remove water from a ﬂooded basement, two pumps, each rated at 40 gallons per minute, are used. After half an hour, the one pump burns out, and the second pump ﬁnishes removing the water half an hour later. How many gallons of water were removed from the basement? 24. A faucet can ﬁll a sink in 5 minutes while a drain will empty the same sink in 8 minutes. If the faucet is turned on and the drain is left open, how long will it take to ﬁll the sink? 25. Working together, Daniel and Donnie can clean the llama pen in 45 minutes. On his own, Daniel can clean the pen in an hour. How long does it take Donnie to clean the llama pen on his own? 354 Rational Functions 26. In Exercise 32, the function C(x) =.03x3 − 4.5x2 + 225x + 250, for x ≥ 0 was used to model the cost (in dollars) to produce x PortaBoy game systems. Using this cost function, ﬁnd the number of PortaBoys which should be produced to minimize the average cost C. Round your answer to the nearest number of systems. 27. Suppose we are in the same situation as Example 4.3.5. If the volume of the box is to be 500 cubic centimeters, use your calculator to ﬁnd the dimensions of the box which minimize the surface area. What is the minimum surface area? Round your
answers to two decimal places. 28. The box for the new Sasquatch-themed cereal, ‘Crypt-Os’, is to have a volume of 140 cubic inches. For aesthetic reasons, the height of the box needs to be 1.62 times the width of the base of the box.13 Find the dimensions of the box which will minimize the surface area of the box. What is the minimum surface area? Round your answers to two decimal places. 29. Sally is Skippy’s neighbor from Exercise 19 in Section 2.3. Sally also wants to plant a vegetable garden along the side of her home. She doesn’t have any fencing, but wants to keep the size of the garden to 100 square feet. What are the dimensions of the garden which will minimize the amount of fencing she needs to buy? What is the minimum amount of fencing she needs to buy? Round your answers to the nearest foot. (Note: Since one side of the garden will border the house, Sally doesn’t need fencing along that side.) 30. Another Classic Problem: A can is made in the shape of a right circular cylinder and is to hold one pint. (For dry goods, one pint is equal to 33.6 cubic inches.)14 (a) Find an expression for the volume V of the can in terms of the height h and the base radius r. (b) Find an expression for the surface area S of the can in terms of the height h and the base radius r. (Hint: The top and bottom of the can are circles of radius r and the side of the can is really just a rectangle that has been bent into a cylinder.) (c) Using the fact that V = 33.6, write S as a function of r and state its applied domain. (d) Use your graphing calculator to ﬁnd the dimensions of the can which has minimal surface area. 31. A right cylindrical drum is to hold 7.35 cubic feet of liquid. Find the dimensions (radius of the base and height) of the drum which would minimize the surface area. What is the minimum surface area? Round your answers to two decimal places. 32. In Exercise 71 in Section 1.4, the population of Sasquatch in Portage County was modeled t+15, where t = 0 represents the year 1803. When were there fewer by the function P (t) = 150t than 100 Sasquatch in Portage County? 131
.62 is a crude approximation of the so-called ‘Golden Ratio’ φ = 1+ 2 14According to www.dictionary.com, there are diﬀerent values given for this conversion. We will stick with 33.6in3. 5 √ for this problem. 4.3 Rational Inequalities and Applications 355 In Exercises 33 - 38, translate the following into mathematical equations. 33. At a constant pressure, the temperature T of an ideal gas is directly proportional to its volume V. (This is Charles’s Law) 34. The frequency of a wave f is inversely proportional to the wavelength of the wave λ. 35. The density d of a material is directly proportional to the mass of the object m and inversely proportional to its volume V. 36. The square of the orbital period of a planet P is directly proportional to the cube of the semi-major axis of its orbit a. (This is Kepler’s Third Law of Planetary Motion ) 37. The drag of an object traveling through a ﬂuid D varies jointly with the density of the ﬂuid ρ and the square of the velocity of the object ν. 38. Suppose two electric point charges, one with charge q and one with charge Q, are positioned r units apart. The electrostatic force F exerted on the charges varies directly with the product of the two charges and inversely with the square of the distance between the charges. (This is Coulomb’s Law) 39. According to this webpage, the frequency f of a vibrating string is given by f = T µ where T is the tension, µ is the linear mass15 of the string and L is the length of the vibrating part of the string. Express this relationship using the language of variation. 1 2L 40. According to the Centers for Disease Control and Prevention www.cdc.gov, a person’s Body Mass Index B is directly proportional to his weight W in pounds and inversely proportional to the square of his height h in inches. (a) Express this relationship as a mathematical equation. (b) If a person who was 5 feet, 10 inches tall weighed 235 pounds had a Body Mass Index of 33.7, what is the value of the constant of proportionality? (c) Rewrite the mathematical equation found in part 40a to include the value of the constant found in part 40b and then ﬁnd your Body Mass Index
. 41. We know that the circumference of a circle varies directly with its radius with 2π as the constant of proportionality. (That is, we know C = 2πr.) With the help of your classmates, compile a list of other basic geometric relationships which can be seen as variations. 15Also known as the linear density. It is simply a measure of mass per unit length. 356 4.3.3 Answers 1. x = − 6 7 2. x = 1, x = 2 4. x = −6, x = 2 5. No solution Rational Functions 3. x = −1 6. x = 0, x = ±2 √ 2 7. (−2, ∞) 9. (−1, 0) ∪ (1, ∞) 11. (−∞, −3) ∪ (−3, 2) ∪ (4, ∞) 8. (−2, 3] 10. [0, ∞) 12. −3, − 1 3 ∪ (2, 3) 13. (−1, 0] ∪ (2, ∞) 14. (−∞, −3) ∪ (−2, −1) ∪ (1, ∞) 15. (−∞, 1] ∪ [2, ∞) 17. (−∞, −3) ∪ −2 √ 2, 0 ∪ 2 19. [−3, 0) ∪ (0, 4) ∪ [5, ∞) √ 2, 3 16. (−∞, −6) ∪ (−1, 2) 18. No solution 20. −1, − 1 2 ∪ (1, ∞) 21. 4.5 miles per hour 22. 24 miles per hour 23. 3600 gallons 24. 40 3 ≈ 13.33 minutes 25. 3 hours 26. The absolute minimum of y = C(x) occurs at ≈ (75.73, 59.57). Since x represents the number of game systems, we check C(75) ≈ 59.58 and C(76) ≈ 59.57. Hence, to minimize the average cost, 76 systems should be produced at an average cost of $59.57 per system. 27. The width (and depth) should be 10.00 centimeters, the height should be 5.00 centimeters. The minimum surface area is 300.00 square centimeters. 28. The width of the
base of the box should be ≈ 4.12 inches, the height of the box should be ≈ 6.67 inches, and the depth of the base of the box should be ≈ 5.09 inches; minimum surface area ≈ 164.91 square inches. 29. The dimensions are ≈ 7 feet by ≈ 14 feet; minimum amount of fencing required ≈ 28 feet. 30. (a) V = πr2h (b) S = 2πr2 + 2πrh (c) S(r) = 2πr2 + 67.2 r, Domain r > 0 (d) r ≈ 1.749 in. and h ≈ 3.498 in. 31. The radius of the drum should be ≈ 1.05 feet and the height of the drum should be ≈ 2.12 feet. The minimum surface area of the drum is ≈ 20.93 cubic feet. 32. P (t) < 100 on (−15, 30), and the portion of this which lies in the applied domain is [0, 30). Since t = 0 corresponds to the year 1803, from 1803 through the end of 1832, there were fewer than 100 Sasquatch in Portage County. 4.3 Rational Inequalities and Applications 357 33. T = kV 36. P 2 = ka3 34. 16 f = k λ 37. 17 D = kρν2 39. Rewriting f = 1 2L T µ as f = √ 1 2 L T √ µ 35. d = km V 38. 18 F = kqQ r2 we see that the frequency f varies directly with the square root of the tension and varies inversely with the length and the square root of the linear mass. 40. (a) B = kW h2 (b) 19 k = 702.68 (c) B = 702.68W h2 16The character λ is the lower case Greek letter ‘lambda.’ 17The characters ρ and ν are the lower case Greek letters ‘rho’ and ‘nu,’ respectively. 18Note the similarity to this formula and Newton’s Law of Universal Gravitation as discussed in Example 5. 19The CDC uses 703. 358 Rational Functions Chapter 5 Further Topics in Functions 5.1 Function Composition Before we embark upon any further adventures with functions, we need to take some time to gather our
thoughts and gain some perspective. Chapter 1 ﬁrst introduced us to functions in Section 1.3. At that time, functions were speciﬁc kinds of relations - sets of points in the plane which passed the Vertical Line Test, Theorem 1.1. In Section 1.4, we developed the idea that functions are processes - rules which match inputs to outputs - and this gave rise to the concepts of domain and range. We spoke about how functions could be combined in Section 1.5 using the four basic arithmetic operations, took a more detailed look at their graphs in Section 1.6 and studied how their graphs behaved under certain classes of transformations in Section 1.7. In Chapter 2, we took a closer look at three families of functions: linear functions (Section 2.1), absolute value functions1 (Section 2.2), and quadratic functions (Section 2.3). Linear and quadratic functions were special cases of polynomial functions, which we studied in generality in Chapter 3. Chapter 3 culminated with the Real Factorization Theorem, Theorem 3.16, which says that all polynomial functions with real coeﬃcients can be thought of as products of linear and quadratic functions. Our next step was to enlarge our ﬁeld2 of study to rational functions in Chapter 4. Being quotients of polynomials, we can ultimately view this family of functions as being built up of linear and quadratic functions as well. So in some sense, Chapters 2, 3, and 4 can be thought of as an exhaustive study of linear and quadratic3 functions and their arithmetic combinations as described in Section 1.5. We now wish x and g(x) = x2/3, and the purpose of the to study other algebraic functions, such as f (x) = ﬁrst two sections of this chapter is to see how these kinds of functions arise from polynomial and rational functions. To that end, we ﬁrst study a new way to combine functions as deﬁned below. √ 1These were introduced, as you may recall, as piecewise-deﬁned linear functions. 2This is a really bad math pun. 3If we broaden our concept of functions to allow for complex valued coeﬃcients, the Complex Factorization Theorem, Theorem 3.14, tells us every function we have studied thus
far is a combination of linear functions. 360 Further Topics in Functions Deﬁnition 5.1. Suppose f and g are two functions. The composite of g with f, denoted g ◦ f, is deﬁned by the formula (g ◦ f )(x) = g(f (x)), provided x is an element of the domain of f and f (x) is an element of the domain of g. The quantity g ◦ f is also read ‘g composed with f ’ or, more simply ‘g of f.’ At its most basic level, Deﬁnition 5.1 tells us to obtain the formula for (g ◦ f ) (x), we replace every occurrence of x in the formula for g(x) with the formula we have for f (x). If we take a step back and look at this from a procedural, ‘inputs and outputs’ perspective, Deﬁntion 5.1 tells us the output from g ◦ f is found by taking the output from f, f (x), and then making that the input to g. The result, g(f (x)), is the output from g ◦ f. From this perspective, we see g ◦ f as a two step process taking an input x and ﬁrst applying the procedure f then applying the procedure g. Abstractly, we have f g x f (x) g ◦ f g(f (x)) In the expression g(f (x)), the function f is often called the ‘inside’ function while g is often called the ‘outside’ function. There are two ways to go about evaluating composite functions - ‘inside out’ and ‘outside in’ - depending on which function we replace with its formula ﬁrst. Both ways are demonstrated in the following example. Example 5.1.1. Let f (x) = x2 − 4x, g(x) = 2 − In numbers 1 - 3, ﬁnd the indicated function value. √ x + 3, and h(x) = 2x x + 1. 1. (g ◦ f )(1) 2. (f ◦ g)(1) 3. (g ◦ g)(6) In numbers 4 - 10, ﬁnd and simplify the indicated composite functions. State the domain
of each. 4. (g ◦ f )(x) 5. (f ◦ g)(x) 6. (g ◦ h)(x) 7. (h ◦ g)(x) 8. (h ◦ h)(x) 9. (h ◦ (g ◦ f ))(x) 10. ((h ◦ g) ◦ f )(x) Solution. 1. Using Deﬁnition 5.1, (g ◦ f )(1) = g(f (1)). We ﬁnd f (1) = −3, so (g ◦ f )(1) = g(f (1)) = g(−3) = 2 5.1 Function Composition 361 2. As before, we use Deﬁnition 5.1 to write (f ◦ g)(1) = f (g(1)). We ﬁnd g(1) = 0, so (f ◦ g)(1) = f (g(1)) = f (0) = 0 3. Once more, Deﬁnition 5.1 tells us (g ◦ g)(6) = g(g(6)). That is, we evaluate g at 6, then plug that result back into g. Since g(6) = −1, (g ◦ g)(6) = g(g(6)) = g(−1) = 2 − √ 2 4. By deﬁnition, (g ◦ f )(x) = g(f (x)). We now illustrate two ways to approach this problem. inside out: We insert the expression f (x) into g ﬁrst to get (g ◦ f )(x) = g(f (x)) = g x2 − 4x = 2 − √ Hence, (g ◦ f )(x) = 2 − x2 − 4x + 3. (x2 − 4x) + 3 = 2 − x2 − 4x + 3 outside in: We use the formula for g ﬁrst to get (g ◦ f )(x) = g(f (x)) = 2 − f (x) + 3 = 2 − (x2 − 4x) + 3 = 2 − x2 − 4x + 3 We get the same answer as before, (g �
� f )(x) = 2 − √ x2 − 4x + 3. To ﬁnd the domain of g ◦ f, we need to ﬁnd the elements in the domain of f whose outputs f (x) are in the domain of g. We accomplish this by following the rule set forth in Section 1.4, that is, we ﬁnd the domain before we simplify. To that end, we examine (g ◦ f )(x) = 2 − (x2 − 4x) + 3. To keep the square root happy, we solve the inequality x2 − 4x + 3 ≥ 0 by creating a sign diagram. If we let r(x) = x2 − 4x + 3, we ﬁnd the zeros of r to be x = 1 and x = 3. We obtain (+) 0 (−) 0 (+) 1 3 Our solution to x2 − 4x + 3 ≥ 0, and hence the domain of g ◦ f, is (−∞, 1] ∪ [3, ∞). 5. To ﬁnd (f ◦ g)(x), we ﬁnd f (g(x)). inside out: We insert the expression g(x) into f ﬁrst to get √ (f ◦ g)(x) = f (g(x)) = + 32 − + 32 − 8 + 4 √ x + 3 362 Further Topics in Functions outside in: We use the formula for f (x) ﬁrst to get (f ◦ g)(x) = f (g(x)) = (g(x))2 − 4 (g(x)) √ √ x + 32 − same algebra as before Thus we get (f ◦ g)(x) = x − 1. To ﬁnd the domain of (f ◦ g), we look to the step before we did any simpliﬁcation and ﬁnd (f ◦ g)(x) = 2 − x + 3. To keep the square root happy, we set x + 3 ≥ 0 and ﬁnd our domain to be [−3, ∞). x + 32 − 4 2 − √ √ 6. To ﬁnd (g ◦ h)(x), we compute g(h(x)). inside out: We insert the expression h(x
) into g ﬁrst to get (g ◦ h)(x) = g(h(x)) = g 2x − 2x x + 1 + 3 3(x + 1) x + 1 get common denominators + 2x x + 1 5x + 3 x + 1 outside in: We use the formula for g(x) ﬁrst to get (g ◦ h)(x) = g(h(x)) = 2 − h(x) + 3 = 2 − = 2 − 2x x + 1 5x + 3 x + 1 + 3 get common denominators as before To ﬁnd the domain of (g ◦ h), we look to the step before we began to simplify: (g ◦ h)(x) = 2 − 2x x + 1 + 3 To avoid division by zero, we need x = −1. To keep the radical happy, we need to solve 2x x + 1 + 3 = 5x + 3 x + 1 ≥ 0 Deﬁning r(x) = 5x+3 x+1, we see r is undeﬁned at x = −1 and r(x) = 0 at x = − 3 5. We get 5.1 Function Composition 363 (+) (−) 0 (+) −1 − 3 5 Our domain is (−∞, −1) ∪ − 3 5, ∞. 7. We ﬁnd (h ◦ g)(x) by ﬁnding h(g(x)). inside out: We insert the expression g(x) into h ﬁrst to get x + 3 √ √ (h ◦ g)(x) = h(g(x)) = − √ √ = = outside in: We use the formula for h(x) ﬁrst to get (h ◦ g)(x) = h(g(x)) = √ 2 (g(x)) (g(x)) + − √ √ = = To ﬁnd the domain of h ◦ g, we look to the step before any simpliﬁcation: (h ◦ g)(x √ To keep the square root happy, we require x+3 ≥ 0 or x ≥ −3. Setting the denominator equal to zero gives 2 − x + 3 + 1 = 0 or x + 3
= 3. Squaring both sides gives us x + 3 = 9, or x = 6. Since x = 6 checks in the original equation, we know x = 6 is the only zero of the denominator. Hence, the domain of h ◦ g is [−3, 6) ∪ (6, ∞). √ √ 8. To ﬁnd (h ◦ h)(x), we substitute the function h into itself, h(h(x)). inside out: We insert the expression h(x) into h to get (h ◦ h)(x) = h(h(x)) = h 2x x + 1 364 Further Topics in Functions 2x x + 1 + 1 2 2x x + 1 4x x + 1 2x x + 1 2x x + 1 4x 1 (x + 1) 2x 1 (x + 1) 4x 3x + 1 · (x + 1) (x + 1) + 1 4x x+1 · (x + 1) · (x + 1) + 1 · (x + 1) · (x + 1) · (x + 1) + x + 1 = = = = = outside in: This approach yields (h ◦ h)(x) = h(h(x)) = 2(h(x)) h(x) + 1 2x x + 1 2 2x x + 1 4x 3x + 1 = = + 1 same algebra as before To ﬁnd the domain of h ◦ h, we analyze (h ◦ h)(x) = 2x x + 1 2 2x x + 1 + 1 To keep the denominator x + 1 happy, we need x = −1. Setting the denominator 2x x + 1 + 1 = 0 gives x = − 1 3. Our domain is (−∞, −1) ∪ −1, − 1 3 ∪ − 1 3, ∞. 5.1 Function Composition 365 9. The expression (h ◦ (g ◦ f ))(x) indicates that we ﬁrst ﬁnd the composite, g ◦ f and compose x2 − 4x + 3. the function h with the result. We know from number 1 that (g ◦ f )(x) = 2 − We now proceed as usual. √ inside out: We insert the expression (g ◦ f
)(x) into h ﬁrst to get (h ◦ (g ◦ f ))(x) = h((g ◦ f )(x)) = h √ x2 − 4x + 3 2 − + 1 x2 − 4x + − √ √ x2 − 4x + 3 x2 − 4x + 3 x2 − 4x + 3 = = outside in: We use the formula for h(x) ﬁrst to get (h ◦ (g ◦ f ))(x) = h((g ◦ f )(x)) = 2 ((g ◦ f )(x)) ((g ◦ f )(x)) + 1 √ x2 − 4x + − √ √ x2 − 4x + 3 x2 − 4x + 3 x2 − 4x + 3 + 1 = = To ﬁnd the domain of (h ◦ (g ◦ f )), we look at the step before we began to simplify, (h ◦ (g ◦ f ))(x) = √ x2 − 4x + 3 2 2 − √ 2 − x2 − 4x + 3 + 1 √ For the square root, we need x2 − 4x + 3 ≥ 0, which we determined in number 1 to be (−∞, 1]∪[3, ∞). Next, we set the denominator to zero and solve: +1 = 0. x2 − 4x + 3 = 3, and, after squaring both sides, we have x2 − 4x + 3 = 9. To solve We get x2 − 4x − 6 = 0, we use the quadratic formula and get x = 2 ± 10. The reader is encouraged x2 − 4x + 3 +1 = 0. to check that both of these numbers satisfy the original equation, Hence we must exclude these numbers from the domain of h ◦ (g ◦ f ). Our ﬁnal domain for h ◦ (f ◦ g) is (−∞, 2 − 10, 1] ∪ 3, 2 + 10 ∪ 2 + x2 − 4x + 3 10, ∞. 10) ∪ (2 − 2 − 2 − √ √ √ √ √ √ √ 10. The expression ((h◦g)◦f
)(x) indicates that we ﬁrst ﬁnd the composite h◦g and then compose that with f. From number 4, we have (h ◦ g)(x 366 Further Topics in Functions We now proceed as before. inside out: We insert the expression f (x) into h ◦ g ﬁrst to get ((h ◦ g) ◦ f )(x) = (h ◦ g)(f (x)) = (h ◦ g) x2 − 4x x2 − 4x) + 3 (x2 − 4x) + 3 x2 − 4x + 3 x2 − 4x + 3 = = outside in: We use the formula for (h ◦ g)(x) ﬁrst to get ((h ◦ g) ◦ f )(x) = (h ◦ g)(f (x)) = 4 − 2 3 − (f (x)) + 3 f (x)) + x2 − 4x) + 3 (x2 − 4x) + 3 x2 − 4x + 3 x2 − 4x + 3 We note that the formula for ((h ◦ g) ◦ f )(x) before simpliﬁcation is identical to that of (h ◦ (g ◦ f ))(x) before we simpliﬁed it. Hence, the two functions have the same domain, h ◦ (f ◦ g) is (−∞, 2 − 10, 1] ∪ 3, 2 + 10 ∪ 2 + 10, ∞. 10) ∪ (2 − √ √ √ √ It should be clear from Example 5.1.1 that, in general, when you compose two functions, such as f and g above, the order matters.4 We found that the functions f ◦ g and g ◦ f were diﬀerent as were g ◦ h and h ◦ g. Thinking of functions as processes, this isn’t all that surprising. If we think of one process as putting on our socks, and the other as putting on our shoes, the order in which we do these two tasks does matter.5 Also note the importance of ﬁnding the domain of the composite function before simplifying. For instance, the domain of f ◦ g is much diﬀerent than its simpl
iﬁed formula would indicate. Composing a function with itself, as in the case of ﬁnding (g ◦ g)(6) and (h ◦ h)(x), may seem odd. Looking at this from a procedural perspective, however, this merely indicates performing a task h and then doing it again - like setting the washing machine to do a ‘double rinse’. Composing a function with itself is called ‘iterating’ the function, and we could easily spend an entire course on just that. The last two problems in Example 5.1.1 serve to demonstrate the associative property of functions. That is, when composing three (or more) functions, as long as we keep the order the same, it doesn’t matter which two functions we compose ﬁrst. This property as well as another important property are listed in the theorem below. 4This shows us function composition isn’t commutative. An example of an operation we perform on two functions which is commutative is function addition, which we deﬁned in Section 1.5. In other words, the functions f + g and g + f are always equal. Which of the remaining operations on functions we have discussed are commutative? 5A more mathematical example in which the order of two processes matters can be found in Section 1.7. In fact, all of the transformations in that section can be viewed in terms of composing functions with linear functions. 5.1 Function Composition 367 Theorem 5.1. Properties of Function Composition: Suppose f, g, and h are functions. h ◦ (g ◦ f ) = (h ◦ g) ◦ f, provided the composite functions are deﬁned. If I is deﬁned as I(x) = x for all real numbers x, then I ◦ f = f ◦ I = f. By repeated applications of Deﬁnition 5.1, we ﬁnd (h ◦ (g ◦ f ))(x) = h((g ◦ f )(x)) = h(g(f (x))). Similarly, ((h ◦ g) ◦ f )(x) = (h ◦ g)(f (x)) = h(g(f (x))). This establishes that the formulas for the two functions are the same. We leave it to
the reader to think about why the domains of these two functions are identical, too. These two facts establish the equality h ◦ (g ◦ f ) = (h ◦ g) ◦ f. A consequence of the associativity of function composition is that there is no need for parentheses when we write h ◦ g ◦ f. The second property can also be veriﬁed using Deﬁnition 5.1. Recall that the function I(x) = x is called the identity function and was introduced in Exercise 73 in Section 2.1. If we compose the function I with a function f, then we have (I ◦ f )(x) = I(f (x)) = f (x), and a similar computation shows (f ◦ I)(x) = f (x). This establishes that we have an identity for function composition much in the same way the real number 1 is an identity for real number multiplication. That is, just as for any real number x, we have for any function. We shall see the concept of an identity take on great signiﬁcance in the next section. Out in the wild, function composition is often used to relate two quantities which may not be directly related, but have a variable in common, as illustrated in our next example. Example 5.1.2. The surface area S of a sphere is a function of its radius r and is given by the formula S(r) = 4πr2. Suppose the sphere is being inﬂated so that the radius of the sphere is increasing according to the formula r(t) = 3t2, where t is measured in seconds, t ≥ 0, and r is measured in inches. Find and interpret (S ◦ r)(t). Solution. If we look at the functions S(r) and r(t) individually, we see the former gives the surface area of a sphere of a given radius while the latter gives the radius at a given time. So, given a speciﬁc time, t, we could ﬁnd the radius at that time, r(t) and feed that into S(r) to ﬁnd the surface area at that time. From this we see that the surface area S is ultimately a function of time t and we ﬁnd (S ◦ r)(t) = S(r(t)) = 4π(r
(t))2 = 4π 3t22 = 36πt4. This formula allows us to compute the surface area directly given the time without going through the ‘middle man’ r. A useful skill in Calculus is to be able to take a complicated function and break it down into a composition of easier functions which our last example illustrates. Example 5.1.3. Write each of the following functions as a composition of two or more (nonidentity) functions. Check your answer by performing the function composition. 1. F (x) = \|3x − 1\| 2. G(x) = 2 x2 + 1 3. H(x Solution. There are many approaches to this kind of problem, and we showcase a diﬀerent methodology in each of the solutions below. 368 Further Topics in Functions 1. Our goal is to express the function F as F = g ◦ f for functions g and f. From Deﬁnition 5.1, we know F (x) = g(f (x)), and we can think of f (x) as being the ‘inside’ function and g as being the ‘outside’ function. Looking at F (x) = \|3x − 1\| from an ‘inside versus outside’ perspective, we can think of 3x − 1 being inside the absolute value symbols. Taking this cue, we deﬁne f (x) = 3x − 1. At this point, we have F (x) = \|f (x)\|. What is the outside function? The function which takes the absolute value of its input, g(x) = \|x\|. Sure enough, (g ◦ f )(x) = g(f (x)) = \|f (x)\| = \|3x − 1\| = F (x), so we are done. 2. We attack deconstructing G from an operational approach. Given an input x, the ﬁrst step is to square x, then add 1, then divide the result into 2. We will assign each of these steps a function so as to write G as a composite of three functions: f, g and h. Our ﬁrst function, f, is the function that squares its input, f (x) = x2. The next function is the function that adds 1 to its input, g(x) = x + 1. Our last
function takes its input and divides it into 2, h(x) = 2 x. The claim is that G = h ◦ g ◦ f. We ﬁnd (h ◦ g ◦ f )(x) = h(g(f (x))) = h(g x2) = h x2 + 1 = 2 x2 + 1 = G(x), so we are done. 3. If we look H(x) = √ x+1√ x−1 with an eye towards building a complicated function from simpler If we deﬁne x is a simple piece of the larger function. f (x)−1. If we want to decompose H = g ◦ f, then we can glean functions, we see the expression f (x) = the formula for g(x) by looking at what is being done to f (x). We take g(x) = x+1 x, we have H(x) = f (x)+1 √ √ x−1, so (g ◦ f )(x) = g(f (x)) = f (x) + 1 f (x(x), as required. 5.1 Function Composition 369 5.1.1 Exercises In Exercises 1 - 12, use the given pair of functions to ﬁnd the following values if they exist. (g ◦ f )(0) (g ◦ f )(−3) (f ◦ g)(−1) (f ◦ g) 1 2 (f ◦ f )(2) (f ◦ f )(−2) 1. f (x) = x2, g(x) = 2x + 1 2. f (x) = 4 − x, g(x) = 1 − x2 3. f (x) = 4 − 3x, g(x) = \|x\| 5. f (x) = 4x + 5, g(x) = √ x 4. f (x) = \|x − 1\|, g(x) = x2 − 5 √ 6. f (x) = 3 − x, g(x) = x2 + 1 7. f (x) = 6 − x − x2, g(x) = x √ x + 10 √ 8. f (x) = 3 x + 1,
g(x) = 4x2 − x 9. f (x) = 3 1 − x, g(x) = 4x x2 + 1 11. f (x) = 2x 5 − x2, g(x) = √ 4x + 1 10. f (x) = x x + 5, g(x) = 2 7 − x2 12. f (x) = √ 2x + 5, g(x) = 10x x2 + 1 In Exercises 13 - 24, use the given pair of functions to ﬁnd and simplify expressions for the following functions and state the domain of each using interval notation. (g ◦ f )(x) (f ◦ g)(x) (f ◦ f )(x) 13. f (x) = 2x + 3, g(x) = x2 − 9 14. f (x) = x2 − x + 1, g(x) = 3x − 5 15. f (x) = x2 − 4, g(x) = \|x\| 16. f (x) = 3x − 5, g(x) = 17. f (x) = \|x + 1\|, g(x) = √ x 19. f (x) = \|x\|, g(x) = √ 4 − x 21. f (x) = 3x − 1, g(x) = 1 x + 3 23. f (x) = x 2x + 1, g(x) = 2x + 1 x √ x √ x + 1 18. f (x) = 3 − x2, g(x) = 20. f (x) = x2 − x − 1, g(x) = √ x − 5 22. f (x) = 3x x − 1, g(x) = x x − 3 24. f (x) = 2x x2 − 4, g(x) = √ 1 − x 370 Further Topics in Functions In Exercises 25 - 30, use f (x) = −2x, g(x) = for the following functions and state the domain of each using interval notation. x and h(x) = \|x\| to ﬁnd and simplify expressions √ 25. (h ◦ g ◦ f )(x) 26. (h ◦
f ◦ g)(x) 27. (g ◦ f ◦ h)(x) 28. (g ◦ h ◦ f )(x) 29. (f ◦ h ◦ g)(x) 30. (f ◦ g ◦ h)(x) In Exercises 31 - 40, write the given function as a composition of two or more non-identity functions. (There are several correct answers, so check your answer using function composition.) 31. p(x) = (2x + 3)3 33. h(x) = √ 2x − 1 35. r(x) = 37. q(x) = 39. v(x) = 2 5x + 1 \|x\| + 1 \|x\| − 1 2x + 1 3 − 4x 32. P (x) = x2 − x + 15 34. H(x) = \|7 − 3x\| 36. R(x) = 38. Q(x) = 40. w(x) = 7 x2 − 1 2x3 + 1 x3 − 1 x2 x4 + 1 41. Write the function F (x) = x3 + 6 x3 − 9 as a composition of three or more non-identity functions. 42. Let g(x) = −x, h(x) = x + 2, j(x) = 3x and k(x) = x − 4. In what order must these functions √ √ be composed with f (x) = x to create F (x) = 3 −x + 2 − 4? 43. What linear functions could be used to transform f (x) = x3 into F (x) = − 1 2 (2x − 7)3 + 1? What is the proper order of composition? In Exercises 44 - 55, let f be the function deﬁned by f = {(−3, 4), (−2, 2), (−1, 0), (0, 1), (1, 3), (2, 4), (3, −1)} and let g be the function deﬁned g = {(−3, −2), (−2, 0), (−1, −4), (0, 0), (1, −3), (2, 1), (3, 2)}. Find the value if it exists. 44. (f ◦ g
)(3) 47. (f ◦ g)(−3) 45. f (g(−1)) 48. (g ◦ f )(3) 46. (f ◦ f )(0) 49. g(f (−3)) 5.1 Function Composition 371 50. (g ◦ g)(−2) 51. (g ◦ f )(−2) 52. g(f (g(0))) 53. f (f (f (−1))) 54. f (f (f (f (f (1))))) 55. (g ◦ g ◦ · · · ◦ g) n times (0) In Exercises 56 - 61, use the graphs of y = f (x) and y = g(x) below to ﬁnd the function valuex) 56. (g ◦ f )(1) 59. (f ◦ g)(0) 57. (f ◦ g)(3) 60. (f ◦ f )(1) y = g(x) 58. (g ◦ f )(2) 61. (g ◦ g)(1) 62. The volume V of a cube is a function of its side length x. Let’s assume that x = t + 1 is also a function of time t, where x is measured in inches and t is measured in minutes. Find a formula for V as a function of t. 63. Suppose a local vendor charges $2 per hot dog and that the number of hot dogs sold per hour x is given by x(t) = −4t2 + 20t + 92, where t is the number of hours since 10 AM, 0 ≤ t ≤ 4. (a) Find an expression for the revenue per hour R as a function of x. (b) Find and simplify (R ◦ x) (t). What does this represent? (c) What is the revenue per hour at noon? 64. Discuss with your classmates how ‘real-world’ processes such as ﬁlling out federal income tax forms or computing your ﬁnal course grade could be viewed as a use of function composition. Find a process for which composition with itself (iteration) makes sense. 372 5.1.2 Answers 1. For f (x) = x2 and g(x) = 2x + 1, Further Topics in Functions (g ◦ f )
(0) = 1 (f ◦ g)(−1) = 1 (f ◦ f )(2) = 16 (g ◦ f )(−3) = 19 (f ◦ g) 1 2 = 4 (f ◦ f )(−2) = 16 2. For f (x) = 4 − x and g(x) = 1 − x2, (g ◦ f )(0) = −15 (f ◦ g)(−1) = 4 (f ◦ f )(2) = 2 (g ◦ f )(−3) = −48 (f ◦ g) 1 2 = 13 4 (f ◦ f )(−2) = −2 3. For f (x) = 4 − 3x and g(x) = \|x\|, (g ◦ f )(0) = 4 (f ◦ g)(−1) = 1 (f ◦ f )(2) = 10 (g ◦ f )(−3) = 13 (f ◦ g) 1 2 = 5 2 (f ◦ f )(−2) = −26 4. For f (x) = \|x − 1\| and g(x) = x2 − 5, (g ◦ f )(0) = −4 (f ◦ g)(−1) = 5 (f ◦ f )(2) = 0 (g ◦ f )(−3) = 11 (f ◦ g) 1 2 = 23 4 (f ◦ f )(−2) = 2 5. For f (x) = 4x + 5 and g(x) = √ x, (g ◦ f )(0) = √ 5 (f ◦ g)(−1) is not real (f ◦ f )(2) = 57 (g ◦ f )(−3) is not real (f ◦ gf ◦ f )(−2) = −7 6. For f (x) = √ 3 − x and g(x) = x2 + 1, (g ◦ f )(0) = 4 (g ◦ f )(−3) = 7 (f ◦ g)(−1) = 1 (f ◦ g) 1 2 = √ 7 2 (f ◦ f )(2) = √ 2 (
f ◦ f )(−2) = 3 − √ 5 5.1 Function Composition 373 7. For f (x) = 6 − x − x2 and g(x) = x √ x + 10, (g ◦ f )(0) = 24 (f ◦ g)(−1) = 0 (f ◦ f )(2) = 6 (g ◦ f )(−3) = 0 (f ◦ g) 1 2 √ = 27−2 8 42 (f ◦ f )(−2) = −14 √ 8. For f (x) = 3 x + 1 and g(x) = 4x2 − x, (g ◦ f )(0) = 3 √ (g ◦ f )(−3f ◦ g)(−1) = 3 6 (f ◦ g) 1 2 = 3√ 12 2 √ (f ◦ f )(2) = 3 3 3 + 1 (f ◦ f )(−2) = 0 9. For f (x) = 3 1−x and g(x) = 4x x2+1, (g ◦ f )(0) = 6 5 (g ◦ f )(−3) = 48 25 (f ◦ g)(−1) = 1 (f ◦ g) 1 2 = −5 (f ◦ f )(2) = 3 4 (f ◦ f )(−2) is undeﬁned 10. For f (x) = x x+5 and g(x) = 2 7−x2, (g ◦ f )(0) = 2 7 (g ◦ f )(−3) = 8 19 (f ◦ g)(−1) = 1 16 = 8 143 (f ◦ g) 1 2 (f ◦ f )(2) = 2 37 (f ◦ f )(−2) = − 2 13 11. For f (x) = 2x 5−x2 and g(x) = √ 4x + 1, (g ◦ f )(0) = 1 (g ◦ f )(−3) = √ 7 (f ◦ g)(−1) is not real (f ◦ g) 1 2 = √ 3 (f ◦ f )(2) = − 8
11 (f ◦ f )(−2) = 8 11 12. For f (x) = √ 2x + 5 and g(x) = 10x x2+1, (g ◦ f )(0) = 5 √ 3 5 (f ◦ g)(−1) is not real (f ◦ f )(2) = √ 11 (g ◦ f )(−3) is not real (f ◦ g) 1 2 = √ 13 (f ◦ f )(−2) = √ 7 13. For f (x) = 2x + 3 and g(x) = x2 − 9 (g ◦ f )(x) = 4x2 + 12x, domain: (−∞, ∞) (f ◦ g)(x) = 2x2 − 15, domain: (−∞, ∞) (f ◦ f )(x) = 4x + 9, domain: (−∞, ∞) 374 Further Topics in Functions 14. For f (x) = x2 − x + 1 and g(x) = 3x − 5 (g ◦ f )(x) = 3x2 − 3x − 2, domain: (−∞, ∞) (f ◦ g)(x) = 9x2 − 33x + 31, domain: (−∞, ∞) (f ◦ f )(x) = x4 − 2x3 + 2x2 − x + 1, domain: (−∞, ∞) 15. For f (x) = x2 − 4 and g(x) = \|x\| (g ◦ f )(x) = \|x2 − 4\|, domain: (−∞, ∞) (f ◦ g)(x) = \|x\|2 − 4 = x2 − 4, domain: (−∞, ∞) (f ◦ f )(x) = x4 − 8x2 + 12, domain: (−∞, ∞) 16. For f (x) = 3x − 5 and g(x) = √ x (g ◦ f )(x) = (f ◦ g)(x) = 3 √ 3x − 5, domain: 5 3, ∞ √ x − 5, domain: [0, ∞) (f �
� f )(x) = 9x − 20, domain: (−∞, ∞) 17. For f (x) = \|x + 1\| and g(x) = √ x (g ◦ f )(x) = \|x + 1\|, domain: (−∞, ∞) √ √ (f ◦ g)(x) = \| x + 1\| = x + 1, domain: [0, ∞) (f ◦ f )(x) = \|\|x + 1\| + 1\| = \|x + 1\| + 1, domain: (−∞, ∞) 18. For f (x) = 3 − x2 and g(x) = √ x + 1 (g ◦ f )(x) = √ 4 − x2, domain: [−2, 2] (f ◦ g)(x) = 2 − x, domain: [−1, ∞) (f ◦ f )(x) = −x4 + 6x2 − 6, domain: (−∞, ∞) 19. For f (x) = \|x\| and g(x) = √ 4 − x (g ◦ f )(x) = 4 − \|x\|, domain: [−4, 4] √ √ (f ◦ g)(x) = \| 4 − x\| = 4 − x, domain: (−∞, 4] (f ◦ f )(x) = \|\|x\|\| = \|x\|, domain: (−∞, ∞) 5.1 Function Composition 375 20. For f (x) = x2 − x − 1 and g(x) = √ x − 5 √ (g ◦ f )(x) = (f ◦ g)(x) = x − 6 − (f ◦ f )(x) = x4 − 2x3 − 2x2 + 3x + 1, domain: (−∞, ∞) x2 − x − 6, domain: (−∞, −2] ∪ [3, ∞) x − 5, domain: [5, ∞) √ 21. For f (x) = 3x − 1 and g(x) = 1 x+3 (g ◦ f )(x) = 1 (f ◦ g)(x) = −
x (f ◦ f )(x) = 9x − 4, domain: (−∞, ∞) 3, ∞ 3x+2, domain: −∞, − 2 x+3, domain: (−∞, −3) ∪ (−3, ∞) ∪ − 2 3 22. For f (x) = 3x x−1 and g(x) = x x−3 (g ◦ f )(x) = x, domain: (−∞, 1) ∪ (1, ∞) (f ◦ g)(x) = x, domain: (−∞, 3) ∪ (3, ∞) (f ◦ f )(x) = 9x 2x+1, domain: −∞, − 1, ∞) 23. For f (x) = x 2x+1 and g(x) = 2x+1 x (g ◦ f )(x) = 4x+1 (f ◦ g)(x) = 2x+1 (f ◦ f )(x) = x 2 x, domain: −∞, − 1 5x+2, domain: −∞, − 2 4x+1, domain: −∞, − 1 √ 5 2 24. For f (x) = 2x x2−4 and g(x), ∪(0, ∞ 5, 0 ∪ (0, ∞) ∪ − g ◦ f )(x) = x2−2x−4 x2−4, domain: (−∞, −2) ∪ 1 − √ 5, 2 ∪ 1 + √ 5, ∞ (f ◦ g)(x) = − 2 √ 1−x x+3, domain: (−∞, −3) ∪ (−3, 1] 4x−x3 17 √ (f ◦ f )(x) = √ √ 1− 17, −1+ 2 2 x4−9x2+16, domain: √ 17 ∪ −1+ 2, 2 ∪ 17 −∞, − 1+ √ 17 2 ∪ 2, 1+ 2 − 1+ 17, ∞ ∪ √ 2 1+ √ 2 17, −2 ∪ −2, 1− √ 2 17 ∪
25. (h ◦ g ◦ f )(x) = \| √ 26. (h ◦ f ◦ g)(x) = \| − 2 √ −2x\| = √ −2x, domain: (−∞, 0] √ x, domain: [0, ∞) 27. (g ◦ f ◦ h)(x) = 28. (g ◦ h ◦ f )(x) = 29. (f ◦ h ◦ g)(x) = −2\| x\| = −2 x\| = 2 −2\|x\|, domain: {0} \| − 2x\| = √ 2\|x\|, domain: (−∞, ∞) √ x, domain: [0, ∞) 376 Further Topics in Functions 30. (f ◦ g ◦ h)(x) = −2 \|x\|,, domain: (−∞, ∞) 31. Let f (x) = 2x + 3 and g(x) = x3, then p(x) = (g ◦ f )(x). 32. Let f (x) = x2 − x + 1 and g(x) = x5, P (x) = (g ◦ f )(x). 33. Let f (x) = 2x − 1 and g(x) = √ x, then h(x) = (g ◦ f )(x). 34. Let f (x) = 7 − 3x and g(x) = \|x\|, then H(x) = (g ◦ f )(x). 35. Let f (x) = 5x + 1 and g(x) = 2 x, then r(x) = (g ◦ f )(x). 36. Let f (x) = x2 − 1 and g(x) = 7 x, then R(x) = (g ◦ f )(x). 37. Let f (x) = \|x\| and g(x) = x+1 x−1, then q(x) = (g ◦ f )(x). 38. Let f (x) = x3 and g(x) = 2x+1 x−1, then Q(x) = (g ◦ f )(x). 39. Let f (x) = 2x and
g(x) = x+1 3−2x, then v(x) = (g ◦ f )(x). 40. Let f (x) = x2 and g(x) = x x2+1, then w(x) = (g ◦ f )(x). 41. F (x) = x3+6 x3−9 = (h(g(f (x))) where f (x) = x3, g(x) = x+6 √ 42. F (x) = 3 −x + 2 − 4 = k(j(f (h(g(x))))) x−9 and h(x) = √ x. 43. One possible solution is F (x) = − 1 2 (2x − 7)3 + 1 = k(j(f (h(g(x))))) where g(x) = 2x, h(x) = 2 x and k(x) = x + 1. You could also have F (x) = H(f (G(x))) where x − 7, j(x) = − 1 G(x) = 2x − 7 and H(x) = − 1 2 x + 1. 44. (f ◦ g)(3) = f (g(3)) = f (2) = 4 45. f (g(−1)) = f (−4) which is undeﬁned 46. (f ◦ f )(0) = f (f (0)) = f (1) = 3 47. (f ◦ g)(−3) = f (g(−3)) = f (−2) = 2 48. (g ◦ f )(3) = g(f (3)) = g(−1) = −4 49. g(f (−3)) = g(4) which is undeﬁned 50. (g ◦ g)(−2) = g(g(−2)) = g(0) = 0 51. (g ◦ f )(−2) = g(f (−2)) = g(2) = 1 52. g(f (g(0))) = g(f (0)) = g(1) = −3 53. f (f (f (−1))) = f (f (0)) = f (1) = 3 54. f
(f (f (f (f (1))))) = f (f (f (f (3)))) = f (f (f (−1))) = f (f (0)) = f (1) = 3 55. (g ◦ g ◦ · · · ◦ g) (0) = 0 n times 5.1 Function Composition 377 56. (g ◦ f )(1) = 3 57. (f ◦ g)(3) = 4 58. (g ◦ f )(2) = 0 59. (f ◦ g)(0) = 4 60. (f ◦ f )(1) = 3 61. (g ◦ g)(1) = 0 62. V (x) = x3 so V (x(t)) = (t + 1)3 63. (a) R(x) = 2x (b) (R ◦ x) (t) = −8t2 + 40t + 184, 0 ≤ t ≤ 4. This gives the revenue per hour as a function of time. (c) Noon corresponds to t = 2, so (R ◦ x) (2) = 232. The hourly revenue at noon is $232 per hour. 378 Further Topics in Functions 5.2 Inverse Functions Thinking of a function as a process like we did in Section 1.4, in this section we seek another function which might reverse that process. As in real life, we will ﬁnd that some processes (like putting on socks and shoes) are reversible while some (like cooking a steak) are not. We start by discussing a very basic function which is reversible, f (x) = 3x + 4. Thinking of f as a process, we start with an input x and apply two steps, as we saw in Section 1.4 1. multiply by 3 2. add 4 To reverse this process, we seek a function g which will undo each of these steps and take the output from f, 3x + 4, and return the input x. If we think of the real-world reversible two-step process of ﬁrst putting on socks then putting on shoes, to reverse the process, we ﬁrst take oﬀ the shoes, and then we take oﬀ the socks. In much the same way, the function g should undo the second step of f ﬁrst. That is, the
function g should 1. subtract 4 2. divide by 3 3 = 15 Following this procedure, we get g(x) = x−4 3. Let’s check to see if the function g does the job. If x = 5, then f (5) = 3(5) + 4 = 15 + 4 = 19. Taking the output 19 from f, we substitute it into g to get g(19) = 19−4 3 = 5, which is our original input to f. To check that g does the job for all x in the domain of f, we take the generic output from f, f (x) = 3x + 4, and substitute that into g. That is, g(f (x)) = g(3x + 4) = (3x+4)−4 3 = x, which is our original input to f. If we carefully examine the arithmetic as we simplify g(f (x)), we actually see g ﬁrst ‘undoing’ the addition of 4, and then ‘undoing’ the multiplication by 3. Not only does g undo f, but f also undoes g. That is, if we take the output from g, g(x) = x−4 3, and put that into f, we get f (g(x)) = f x−4 + 4 = (x − 4) + 4 = x. Using the language of function 3 composition developed in Section 5.1, the statements g(f (x)) = x and f (g(x)) = x can be written as (g ◦ f )(x) = x and (f ◦ g)(x) = x, respectively. Abstractly, we can visualize the relationship between f and g in the diagram below. = 3 x−4 3 = 3x 3 f g y = f (x) x = g(f (x)) 5.2 Inverse Functions 379 The main idea to get from the diagram is that g takes the outputs from f and returns them to their respective inputs, and conversely, f takes outputs from g and returns them to their respective inputs. We now have enough background to state the central deﬁnition of the section. Deﬁnition 5.2. Suppose f and g are two functions such that 1. (g ◦ f )(x) = x for all x in the domain of f and 2
. (f ◦ g)(x) = x for all x in the domain of g then f and g are inverses of each other and the functions f and g are said to be invertible. We now formalize the concept that inverse functions exchange inputs and outputs. Theorem 5.2. Properties of Inverse Functions: Suppose f and g are inverse functions. The rangea of f is the domain of g and the domain of f is the range of g f (a) = b if and only if g(b) = a (a, b) is on the graph of f if and only if (b, a) is on the graph of g aRecall this is the set of all outputs of a function. Theorem 5.2 is a consequence of Deﬁnition 5.2 and the Fundamental Graphing Principle for Functions. We note the third property in Theorem 5.2 tells us that the graphs of inverse functions are reﬂections about the line y = x. For a proof of this, see Example 1.1.7 in Section 1.1 and Exercise 72 in Section 2.1. For example, we plot the inverse functions f (x) = 3x + 4 and g(x) = x−4 3 below. y 2 1 y = f (x) y = x x −2 −1 1 2 −1 −2 y = g(x) If we abstract one step further, we can express the sentiment in Deﬁnition 5.2 by saying that f and g are inverses if and only if g ◦ f = I1 and f ◦ g = I2 where I1 is the identity function restricted1 to the domain of f and I2 is the identity function restricted to the domain of g. In other words, I1(x) = x for all x in the domain of f and I2(x) = x for all x in the domain of g. Using this description of inverses along with the properties of function composition listed in Theorem 5.1, we can show that function inverses are unique.2 Suppose g and h are both inverses of a function 1The identity function I, which was introduced in Section 2.1 and mentioned in Theorem 5.1, has a domain of all real numbers. Since the domains of f and g may not be all real numbers, we need the restrictions listed
here. 2In other words, invertible functions have exactly one inverse. 380 Further Topics in Functions f. By Theorem 5.2, the domain of g is equal to the domain of h, since both are the range of f. This means the identity function I2 applies both to the domain of h and the domain of g. Thus h = h ◦ I2 = h ◦ (f ◦ g) = (h ◦ f ) ◦ g = I1 ◦ g = g, as required.3 We summarize the discussion of the last two paragraphs in the following theorem.4 Theorem 5.3. Uniqueness of Inverse Functions and Their Graphs : Suppose f is an invertible function. There is exactly one inverse function for f, denoted f −1 (read f -inverse) The graph of y = f −1(x) is the reﬂection of the graph of y = f (x) across the line y = x. 3, which is certainly diﬀerent than 1 The notation f −1 is an unfortunate choice since you’ve been programmed since Elementary Algebra to think of this as 1 f. This is most deﬁnitely not the case since, for instance, f (x) = 3x + 4 has as its inverse f −1(x) = x−4 3x+4. Why does this confusing notation persist? As we mentioned in Section 5.1, the identity function I is to function composition what the real number 1 is to real number multiplication. The choice of notation f −1 alludes to the property that f −1 ◦ f = I1 and f ◦ f −1 = I2, in much the same way as 3−1 · 3 = 1 and 3 · 3−1 = 1. Let’s turn our attention to the function f (x) = x2. Is f invertible? A likely candidate for the inverse x2 = \|x\|, which is the function g(x) = is not equal to x for all x in the domain (−∞, ∞). For example, when x = −2, f (−2) = (−2)2 = 4, 4 = 2, which means g failed to return the input −2 from its output 4. What g did, but g(4) = however, is match the output 4 to a diﬀerent
input, namely 2, which satisﬁes f (2) = 4. This issue is presented schematically in the picture below. x. Checking the composition yields (g◦f )(x) = g(f (x)) = f (x2 x = 2 4 We see from the diagram that since both f (−2) and f (2) are 4, it is impossible to construct a function which takes 4 back to both x = 2 and x = −2. (By deﬁnition, a function matches a real number with exactly one other real number.) From a graphical standpoint, we know that if 3It is an excellent exercise to explain each step in this string of equalities. 4In the interests of full disclosure, the authors would like to admit that much of the discussion in the previous paragraphs could have easily been avoided had we appealed to the description of a function as a set of ordered pairs. We make no apology for our discussion from a function composition standpoint, however, since it exposes the reader to more abstract ways of thinking of functions and inverses. We will revisit this concept again in Chapter 8. 5.2 Inverse Functions 381 y = f −1(x) exists, its graph can be obtained by reﬂecting y = x2 about the line y = x, in accordance with Theorem 5.3. Doing so produces y 7 6 5 4 3 2 1 (2, 4) (−2, 4) y (4, 21 −2 −2 −1 1 2 x y = f (x) = x2 reﬂect across y = x −−−−−−−−−−−−−−−→ switch x and y coordinates (4, −2) y = f −1(x)? We see that the line x = 4 intersects the graph of the supposed inverse twice - meaning the graph fails the Vertical Line Test, Theorem 1.1, and as such, does not represent y as a function of x. The vertical line x = 4 on the graph on the right corresponds to the horizontal line y = 4 on the graph of y = f (x). The fact that the horizontal line y = 4 intersects the graph of f twice means two diﬀerent inputs, namely x = −2 and x = 2, are matched with the same output, 4, which is the cause of all of the trouble. In general,
for a function to have an inverse, diﬀerent inputs must go to diﬀerent outputs, or else we will run into the same problem we did with f (x) = x2. We give this property a name. Deﬁnition 5.3. A function f is said to be one-to-one if f matches diﬀerent inputs to diﬀerent outputs. Equivalently, f is one-to-one if and only if whenever f (c) = f (d), then c = d. Graphically, we detect one-to-one functions using the test below. Theorem 5.4. The Horizontal Line Test: A function f is one-to-one if and only if no horizontal line intersects the graph of f more than once. We say that the graph of a function passes the Horizontal Line Test if no horizontal line intersects the graph more than once; otherwise, we say the graph of the function fails the Horizontal Line Test. We have argued that if f is invertible, then f must be one-to-one, otherwise the graph given by reﬂecting the graph of y = f (x) about the line y = x will fail the Vertical Line Test. It turns out that being one-to-one is also enough to guarantee invertibility. To see this, we think of f as the set of ordered pairs which constitute its graph. If switching the x- and y-coordinates of the points results in a function, then f is invertible and we have found f −1. This is precisely what the Horizontal Line Test does for us: it checks to see whether or not a set of points describes x as a function of y. We summarize these results below. 382 Further Topics in Functions Theorem 5.5. Equivalent Conditions for Invertibility: Suppose f is a function. The following statements are equivalent. f is invertible f is one-to-one The graph of f passes the Horizontal Line Test We put this result to work in the next example. Example 5.2.1. Determine if the following functions are one-to-one in two ways: (a) analytically using Deﬁnition 5.3 and (b) graphically using the Horizontal Line Test. 1. f (x) = 1 − 2x 5 3. h(
x) = x2 − 2x + 4 Solution. 2. g(x) = 2x 1 − x 4. F = {(−1, 1), (0, 2), (2, 1)} 1. (a) To determine if f is one-to-one analytically, we assume f (c) = f (d) and attempt to deduce that c = d. f (c) = f (d) 1 − 2c 5 = 1 − 2d 5 1 − 2c = 1 − 2d −2c = −2d c = d Hence, f is one-to-one. (b) To check if f is one-to-one graphically, we look to see if the graph of y = f (x) passes the Horizontal Line Test. We have that f is a non-constant linear function, which means its graph is a non-horizontal line. Thus the graph of f passes the Horizontal Line Test. 2. (a) We begin with the assumption that g(c) = g(d) and try to show c = d. g(c) = g(d) 2d 2c 1 − d 1 − c = 2c(1 − d) = 2d(1 − c) 2c − 2cd = 2d − 2dc 2c = 2d c = d We have shown that g is one-to-one. 5.2 Inverse Functions 383 (b) We can graph g using the six step procedure outlined in Section 4.2. We get the sole intercept at (0, 0), a vertical asymptote x = 1 and a horizontal asymptote (which the graph never crosses) y = −2. We see from that the graph of g passes the Horizontal Line Test. y 2 1 −2 −1 1 2 −1 −2 y = f (x) x −2 −1 y 4 3 2 1 −1 −2 −3 −4 −5 −6 x 1 2 y = g(x) 3. (a) We begin with h(c) = h(d). As we work our way through the problem, we encounter a nonlinear equation. We move the non-zero terms to the left, leave a 0 on the right and factor accordingly. h(c) = h(d) c2 − 2c + 4 = d2 − 2d + 4
c2 − 2c = d2 − 2d c2 − d2 − 2c + 2d = 0 (c + d)(c − d) − 2(c − d) = 0 (c − d)((c + d) − 2) = 0 c − d = 0 or c = d or factor by grouping We get c = d as one possibility, but we also get the possibility that c = 2 − d. This suggests that f may not be one-to-one. Taking d = 0, we get c = 0 or c = 2. With f (0) = 4 and f (2) = 4, we have produced two diﬀerent inputs with the same output meaning f is not one-to-one. (b) We note that h is a quadratic function and we graph y = h(x) using the techniques presented in Section 2.3. The vertex is (1, 3) and the parabola opens upwards. We see immediately from the graph that h is not one-to-one, since there are several horizontal lines which cross the graph more than once. 4. (a) The function F is given to us as a set of ordered pairs. The condition F (c) = F (d) means the outputs from the function (the y-coordinates of the ordered pairs) are the same. We see that the points (−1, 1) and (2, 1) are both elements of F with F (−1) = 1 and F (2) = 1. Since −1 = 2, we have established that F is not one-to-one. (b) Graphically, we see the horizontal line y = 1 crosses the graph more than once. Hence, the graph of F fails the Horizontal Line Test. 384 Further Topics in Functions 1 y = h(x) y 2 1 x −2 −1 1 2 y = F (x) We have shown that the functions f and g in Example 5.2.1 are one-to-one. This means they are invertible, so it is natural to wonder what f −1(x) and g−1(x) would be. For f (x) = 1−2x, we can think our way through the inverse since there is only one occurrence of x. We can track step-by-step what is done to x and reverse those steps as we did at the beginning of
the chapter. The function g(x) = 2x 1−x is a bit trickier since x occurs in two places. When one evaluates g(x) for a speciﬁc value of x, which is ﬁrst, the 2x or the 1 − x? We can imagine functions more complicated than these so we need to develop a general methodology to attack this problem. Theorem 5.2 tells us equation y = f −1(x) is equivalent to f (y) = x and this is the basis of our algorithm. 5 Steps for ﬁnding the Inverse of a One-to-one Function 1. Write y = f (x) 2. Interchange x and y 3. Solve x = f (y) for y to obtain y = f −1(x) Note that we could have simply written ‘Solve x = f (y) for y’ and be done with it. The act of interchanging the x and y is there to remind us that we are ﬁnding the inverse function by switching the inputs and outputs. Example 5.2.2. Find the inverse of the following one-to-one functions. Check your answers analytically using function composition and graphically. 1. f (x) = 1 − 2x 5 Solution. 2. g(x) = 2x 1 − x 1. As we mentioned earlier, it is possible to think our way through the inverse of f by recording the steps we apply to x and the order in which we apply them and then reversing those steps in the reverse order. We encourage the reader to do this. We, on the other hand, will practice the algorithm. We write y = f (x) and proceed to switch x and y 5.2 Inverse Functions 385 y = f (x) y = x = 1 − 2x 5 1 − 2y 5 5x = 1 − 2y switch x and y 5x − 1 = −2y 5x − 1 − We have f −1(x) = − 5 x for all x in the domain of f, which is all real numbers. 2 x+ 1 2. To check this answer analytically, we ﬁrst check that f −1 ◦ f (x) = f −1 ◦ f (x) = f −1(f (x)) 1 f (x) + 2 1 − 2
x 1 − 2x) + + x + 1 2 We now check that f ◦ f −1 (x) = x for all x in the range of f which is also all real numbers. (Recall that the domain of f −1) is the range of f.) f ◦ f −1 (x) = f (f −1(x)) = = = 1 − 2f −1(x + 5x − 1 5 5x = 5 = x To check our answer graphically, we graph y = f (x) and y = f −1(x) on the same set of axes.5 They appear to be reﬂections across the line y = x. 5Note that if you perform your check on a calculator for more sophisticated functions, you’ll need to take advantage of the ‘ZoomSquare’ feature to get the correct geometric perspective. 386 Further Topics in Functions y 2 1 y = x −4 −3 −2 −1 1 2 3 4 x −1 −2 y = f (x) y = f −1(x) 2. To ﬁnd g−1(x), we start with y = g(x). We note that the domain of g is (−∞, 1) ∪ (1, ∞). y = y = g(x) 2x 1 − x 2y 1 − y x = x(1 − y) = 2y x − xy = 2y x = xy + 2y x = y(x + 2) y = x x + 2 switch x and y factor x+2. To check this analytically, we ﬁrst check g−1 ◦ g (x) = x for all x We obtain g−1(x) = x in the domain of g, that is, for all x = 1. g−1 ◦ g (x) = g−1(g(x)) 2x 1 − x = g−1 = = 2x 1 − x 2x 1 − x 2x 1 − x 2x 1 − x + 2 + 2 · (1 − x) (1 − x) clear denominators 5.2 Inverse Functions 387 2x 2x + 2(1 − x) 2x 2x + 2 − 2x = = = 2x 2 = x Next, we check g g−1(x) = x for
all x in the range of g. From the graph of g in Example 5.2.1, we have that the range of g is (−∞, −2) ∪ (−2, ∞). This matches the domain we get from the formula g−1(x) = x x+2, as it should. g ◦ g−1 (x) = g g−1(x 2x (x + 2) − x 2x x + 2) (x + 2) x + 2 clear denominators 1 − = x Graphing y = g(x) and y = g−1(x) on the same set of axes is busy, but we can see the symmetric relationship if we thicken the curve for y = g−1(x). Note that the vertical asymptote x = 1 of the graph of g corresponds to the horizontal asymptote y = 1 of the graph of g−1, as it should since x and y are switched. Similarly, the horizontal asymptote y = −2 of the graph of g corresponds to the vertical asymptote x = −2 of the graph of g−1. 388 Further Topics in Functions 6 −5 −4 −3 −2 −1 1 2 3 4 5 6 x −1 −2 −3 −4 −5 −6 y = g(x) and y = g−1(x) We now return to f (x) = x2. We know that f is not one-to-one, and thus, is not invertible. However, if we restrict the domain of f, we can produce a new function g which is one-to-one. If we deﬁne g(x) = x2, x ≥ 0, then we have 2 −1 1 2 x −2 −1 1 2 x y = f (x) = x2 restrict domain to x ≥ 0 −−−−−−−−−−−−−−−→ y = g(x) = x2, x ≥ 0 The graph of g passes the Horizontal Line Test. To ﬁnd an inverse of g, we proceed as usual y = g(x) y = x2, x ≥ 0 x = y2 switch x and y since y ≥ 0 5.2 Inverse Functions 389 √ We get g−1(x) = x. At ﬁrst
it looks like we’ll run into the same trouble as before, but when we check the composition, the domain restriction on g saves the day. We get g−1 ◦ g (x) = g−1(g(x)) = g−1 x2 = x2 = \|x\| = x, since x ≥ 0. Checking g ◦ g−1 (x) = g g−1(x) = √ x)2 = x. Graphing6 g and g−1 on the same set of axes shows that they are reﬂections g ( about the line y = x. x) = ( √ √ y y = g(x) y = x y = g−1(x Our next example continues the theme of domain restriction. Example 5.2.3. Graph the following functions to show they are one-to-one and ﬁnd their inverses. Check your answers analytically using function composition and graphically. 1. j(x) = x2 − 2x + 4, x ≤ 1. 2. k(x) = √ x + 2 − 1 Solution. 1. The function j is a restriction of the function h from Example 5.2.1. Since the domain of j is restricted to x ≤ 1, we are selecting only the ‘left half’ of the parabola. We see that the graph of j passes the Horizontal Line Test and thus j is invertible1 y = j(x) 6We graphed y = √ x in Section 1.7. 390 Further Topics in Functions We now use our algorithm7 to ﬁnd j−1(x). y = j(x) y = x2 − 2x + 4, x ≤ 1 x = y2 − 2y + 4, y ≤ 1 0 = y2 − 2y + 4 − x 2 ± (−2)2 − 4(1)(4 − x(1) √ 2 ± 4x − 12 2 2 ± 4(x − 3 − switch x and y quadratic formula, c = 4 − x since y ≤ 1. We have j−1(x) = 1 − the domain of j is x ≤ 1. √ x − 3. When we simplify j−1 ◦ j (x), we need to remember that j−1 ◦ j (x) = j−1
(j(x)) (x2 − 2x + 4) − 3 x2 − 2x + 1 (x − 1)2 = j−1 x2 − 2x + x − 1\| = 1 − (−(x − 1)) = x since x ≤ 1 Checking j ◦ j−1, we get j ◦ j−1 (x) = j j−1(x) x − 3 x − 32 − − 32 − 7Here, we use the Quadratic Formula to solve for y. For ‘completeness,’ we note you can (and should!) also consider solving for y by ‘completing’ the square. 5.2 Inverse Functions 391 Using what we know from Section 1.7, we graph y = j−1(x) and y = j(x) below. y = j(x) y 6 5 4 3 2 1 −−1(x) 2. We graph y = k(x) = √ x + 2 − 1 using what we learned in Section 1.7 and see k is one-to-one. y 2 1 −2 −1 1 2 x −1 −2 y = k(x) We now try to ﬁnd k−1. y = k(xx + 1)2 = √ x2 + 2x + + 22 y = x2 + 2x − 1 switch x and y We have k−1(x) = x2 + 2x − 1. Based on our experience, we know something isn’t quite right. We determined k−1 is a quadratic function, and we have seen several times in this section that these are not one-to-one unless their domains are suitably restricted. Theorem 5.2 tells us that the domain of k−1 is the range of k. From the graph of k, we see that the range is [−1, ∞), which means we restrict the domain of k−1 to x ≥ −1. We now check that this works in our compositions. 392 Further Topics in Functions k−1 ◦ k (x) = k−1(k(x)) x + 2 − 1, x ≥ − − 12 x + 22 − − and k ◦ k−1 (x) = k x2 + 2x − 1 x ≥ −1 (x2 + 2x − 1) + 2 − 1 �
� x2 + 2x + 1 − 1 (x + 1)2 − 1 = = = = \|x + 1 since x ≥ −1 Graphically, everything checks out as well, provided that we remember the domain restriction on k−1 means we take the right half of the parabola. y 2 y = k(x) 1 −2 −1 1 2 x −1 y = k−1(x) −2 Our last example of the section gives an application of inverse functions. Example 5.2.4. Recall from Section 2.1 that the price-demand equation for the PortaBoy game system is p(x) = −1.5x + 250 for 0 ≤ x ≤ 166, where x represents the number of systems sold weekly and p is the price per system in dollars. 5.2 Inverse Functions 393 1. Explain why p is one-to-one and ﬁnd a formula for p−1(x). State the restricted domain. 2. Find and interpret p−1(220). 3. Recall from Section 2.3 that the weekly proﬁt P, in dollars, as a result of selling x systems is given by P (x) = −1.5x2 + 170x − 150. Find and interpret P ◦ p−1 (x). 4. Use your answer to part 3 to determine the price per PortaBoy which would yield the maxi- mum proﬁt. Compare with Example 2.3.3. Solution. 1. We leave to the reader to show the graph of p(x) = −1.5x + 250, 0 ≤ x ≤ 166, is a line segment from (0, 250) to (166, 1), and as such passes the Horizontal Line Test. Hence, p is one-to-one. We ﬁnd the expression for p−1(x) as usual and get p−1(x) = 500−2x. The domain of p−1 should match the range of p, which is [1, 250], and as such, we restrict the domain of p−1 to 1 ≤ x ≤ 250. 3 2. We ﬁnd p−1(220) = 500−2(220) = 20. Since the function p took as inputs the weekly sales and furnished the price per system as the output, p−1 takes the price per system and returns
the weekly sales as its output. Hence, p−1(220) = 20 means 20 systems will be sold in a week if the price is set at $220 per system. 3 = −1.5 500−2x 3. We compute P ◦ p−1 (x) = P p−1(x) = P 500−2x − 150. After a hefty amount of Elementary Algebra,8 we obtain P ◦ p−1 (x) = − 2 3 x2 +220x− 40450. To understand what this means, recall that the original proﬁt function P gave us the weekly proﬁt as a function of the weekly sales. The function p−1 gives us the weekly sales as a function of the price. Hence, P ◦ p−1 takes as its input a price. The function p−1 returns the weekly sales, which in turn is fed into P to return the weekly proﬁt. Hence, P ◦ p−1 (x) gives us the weekly proﬁt (in dollars) as a function of the price per system, x, using the weekly sales p−1(x) as the ‘middle man’. + 170 500−2x 2 3 3 3 3 4. We know from Section 2.3 that the graph of y = P ◦ p−1 (x) is a parabola opening downwards. The maximum proﬁt is realized at the vertex. Since we are concerned only with the price per system, we need only ﬁnd the x-coordinate of the vertex. Identifying a = − 2 3 and b = 220, we get, by the Vertex Formula, Equation 2.4, x = − b 2a = 165. Hence, weekly proﬁt is maximized if we set the price at $165 per system. Comparing this with our answer from Example 2.3.3, there is a slight discrepancy to the tune of $0.50. We leave it to the reader to balance the books appropriately. 8It is good review to actually do this! 394 5.2.1 Exercises Further Topics in Functions In Exercises 1 - 20, show that the given function is one-to-one and ﬁnd its inverse. Check your answers algebraically and graphically. Verify that the range of f is the domain of f −
1 and vice-versa. 1. f (x) = 6x − 2 3. f (x) = 5. f (x) = x − 2 3 + 4 √ 3x − 1 + 5 7. f (x. f (x) = 5 3x − 1 2. f (x) = 42 − x 4. f (x) = 1 − 6. f (x) = 2 − 4 + 3x 5 √ x − 5 8. f (x) = 1 − 2 √ 2x + 5 √ 10. f (x) = 3 − 3 x − 2 11. f (x) = x2 − 10x, x ≥ 5 12. f (x) = 3(x + 4)2 − 5, x ≤ −4 13. f (x) = x2 − 6x + 5, x ≤ 3 14. f (x) = 4x2 + 4x + 1, x < −1 15. f (x) = 17. f (x) = 19. f (x) = 3 4 − x 2x − 1 3x + 4 −3x − 2 x + 3 16. f (x) = 18. f (x) = 20. f (x) = x 1 − 3x 4x + 2 3x − 6 x − 2 2x − 1 With help from your classmates, ﬁnd the inverses of the functions in Exercises 21 - 24. 21. f (x) = ax + b, a = 0 22. f (x) = a √ x − h + k, a = 0, x ≥ h 23. f (x) = ax2 + bx + c where a = 0, x ≥ − b 2a. 24. f (x) = ax + b cx + d, (See Exercise 33 below.) 25. In Example 1.5.3, the price of a dOpi media player, in dollars per dOpi, is given as a function of the weekly sales x according to the formula p(x) = 450 − 15x for 0 ≤ x ≤ 30. (a) Find p−1(x) and state its domain. (b) Find and interpret p−1(105). (c) In Example 1.5.3, we determined that the proﬁt (in dollars) made from producing and selling x d
Opis per week is P (x) = −15x2 + 350x − 2000, for 0 ≤ x ≤ 30. Find P ◦ p−1 (x) and determine what price per dOpi would yield the maximum proﬁt. What is the maximum proﬁt? How many dOpis need to be produced and sold to achieve the maximum proﬁt? 5.2 Inverse Functions 395 26. Show that the Fahrenheit to Celsius conversion function found in Exercise 35 in Section 2.1 is invertible and that its inverse is the Celsius to Fahrenheit conversion function. 27. Analytically show that the function f (x) = x3 + 3x + 1 is one-to-one. Since ﬁnding a formula for its inverse is beyond the scope of this textbook, use Theorem 5.2 to help you compute f −1(1), f −1(5), and f −1(−3). 28. Let f (x) = 2x x2−1. Using the techniques in Section 4.2, graph y = f (x). Verify that f is oneto-one on the interval (−1, 1). Use the procedure outlined on Page 384 and your graphing calculator to ﬁnd the formula for f −1(x). Note that since f (0) = 0, it should be the case that f −1(0) = 0. What goes wrong when you attempt to substitute x = 0 into f −1(x)? Discuss with your classmates how this problem arose and possible remedies. 29. With the help of your classmates, explain why a function which is either strictly increasing or strictly decreasing on its entire domain would have to be one-to-one, hence invertible. 30. If f is odd and invertible, prove that f −1 is also odd. 31. Let f and g be invertible functions. With the help of your classmates show that (f ◦ g) is one-to-one, hence invertible, and that (f ◦ g)−1(x) = (g−1 ◦ f −1)(x). 32. What graphical feature must a function f possess for it to be its own inverse? 33. What conditions must you place on the values of a, b, c and d in Exercise 24 in order to guarantee that the function is invertible? 396 Further Topics
in Functions 5.2.2 Answers 1. f −1(x) = x + 2 6 3. f −1(x) = 3x − 10 5. f −1(x) = 1 3 (x − 5)2 + 1 3, x ≥ 5 7. f −1(x) = 1 9 (x + 4)2 + 1, x ≥ −4 9. f −1(x) = 1 3 x5 + 1 √ 3 11. f −1(x) = 5 + x + 25 2. f −1(x) = 42 − x 4. f −1(x. f −1(x) = (x − 2)2 + 5, x ≤ 2 8. f −1(x) = 1 8 (x − 1)2 − 5 2, x ≤ 1 10. f −1(x) = −(x − 3)3 + 2 12. f −1(x) = − x+5 3 − 4 13. f −1(x) = 3 − √ x + 4 15. f −1(x) = 17. f −1(x) = 19. f −1(x) = 4x − 3 x 4x + 1 2 − 3x −3x − 2 x + 3 14. f −1(x) = − √ x+1 2, x > 1 16. f −1(x) = 18. f −1(x) = 20. f −1(x) = x 3x + 1 6x + 2 3x − 4 x − 2 2x − 1 25. 15 15 x2 + 110. The domain of p−1 is the range of p which is [0, 450] (a) p−1(x) = 450−x (b) p−1(105) = 23. This means that if the price is set to $105 then 23 dOpis will be sold. 3 x − 5000, 0 ≤ x ≤ 450. The graph of y = P ◦ p−1 (x) (c) P ◦ p−1 (x) = − 1 is a parabola opening downwards with vertex 275, 125 ≈ (275, 41.67). This means 3 that the maximum proﬁt is a whopping $41.67 when the price per dOpi is set to $275. At this price, we
can produce and sell p−1(275) = 11.6 dOpis. Since we cannot sell part of a system, we need to adjust the price to sell either 11 dOpis or 12 dOpis. We ﬁnd p(11) = 285 and p(12) = 270, which means we set the price per dOpi at either $285 or $270, respectively. The proﬁts at these prices are P ◦ p−1 (285) = 35 and P ◦ p−1 (270) = 40, so it looks as if the maximum proﬁt is $40 and it is made by producing and selling 12 dOpis a week at a price of $270 per dOpi. 27. Given that f (0) = 1, we have f −1(1) = 0. Similarly f −1(5) = 1 and f −1(−3) = −1 5.3 Other Algebraic Functions 397 5.3 Other Algebraic Functions This section serves as a watershed for functions which are combinations of polynomial, and more generally, rational functions, with the operations of radicals. It is business of Calculus to discuss these functions in all the detail they demand so our aim in this section is to help shore up the requisite skills needed so that the reader can answer Calculus’s call when the time comes. We brieﬂy recall the deﬁnition and some of the basic properties of radicals from Intermediate Algebra.1 Deﬁnition 5.4. Let x be a real number and n a natural number.a If n is odd, the principal √ √ nth root of x, denoted n x is the unique real number satisfying ( n x is √ deﬁned similarlyb provided x ≥ 0 and n x ≥ 0. The index is the number n and the radicand √ √ x instead of 2 is the number x. For n = 2, we write √ x)n = x. If n is even, n x. aRecall this means n = 1, 2, 3,.... bRecall both x = −2 and x = 2 satisfy x4 = 16, but 4√ 16 = 2, not −2. √ It is worth remarking that, in light of Section 5.2, we could de
ﬁne f (x) = n x functionally as the inverse of g(x) = xn with the stipulation that when n is even, the domain of g is restricted to [0, ∞). From what we know about g(x) = xn from Section 3.1 along with Theorem 5.3, we can produce √ x by reﬂecting the graphs of g(x) = xn across the line y = x. Below are the the graphs of f (x) = n √ x, y = 4 x. The point (0, 0) is indicated as a reference. The axes are graphs of y = hidden so we can see the vertical steepening near x = 0 and the horizontal ﬂattening as x → ∞. √ x and √ x y = 6√ x The odd-indexed radical functions also follow a predictable trend - steepening near x = 0 and ﬂattening as x → ±∞. In the exercises, you’ll have a chance to graph some basic radical functions using the techniques presented in Section 1.7. y = 3√ x y = 5√ x y = 7√ x We have used all of the following properties at some point in the textbook for the case n = 2 (the square root), but we list them here in generality for completeness. 1Although we discussed imaginary numbers in Section 3.4, we restrict our attention to real numbers in this section. See the epilogue on page 294 for more details. 398 Further Topics in Functions Theorem 5.6. Properties of Radicals: Let x and y be real numbers and m and n be natural √ numbers. If n y are real numbers, then √ x, n √ Product Rule: n √ xy = n √ Powers of Radicals: n Quotient Rule: n x y = √ x n y √ xm = ( n √ n √ n x y x)m, provided y = 0. √ If n is odd, n √ xn = x; if n is even, n xn = \|x\|. √ xy = n The proof of Theorem 5.6 is based on the deﬁnition of the principal roots and properties of expo√ xy nents. To establish the product rule, consider the
following. If n is odd, then by deﬁnition n √ √ xy)n = xy. Given that n is the unique real number such that ( n = xy, √ √ √ it must be the case that n xy is the unique non-negative real x n y. If n is even, then n √ √ xy)n = xy. Also note that since n is even, n number such that ( n y are also non-negative √ √ √ and hence so is n y. The quotient rule is y. Proceeding as above, we ﬁnd that n x n proved similarly and is left as an exercise. The power rule results from repeated application of the √ x is a real number to start with.2 The last property is an application of product rule, so long as n the power rule when n is odd, and the occurrence of the absolute value when n is even is due to √ 16 = 2 = \| − 2\|, not −2. the requirement that n It’s this last property which makes compositions of roots and powers delicate. This is especially true when we use exponential notation for radicals. Recall the following deﬁnition. x ≥ 0 in Deﬁnition 5.4. For instance, 4 √ x and n √ √ xy = n x n √ (−2)4 = 4 √ x)n n √ = ( n √ x n yn yn Deﬁnition 5.5. Let x be a real number, m an integera and n a natural number. √ x and is deﬁned whenever n √ √ m xm, whenever ( n n = ( n x)m is deﬁned. √ x)m = n x is deﬁned. √ 1 n = n x x aRecall this means m = 0, ±1, ±2,... The rational exponents deﬁned in Deﬁnition 5.5 behave very similarly to the usual integer exponents from Elementary Algebra with one critical exception. Consider the expression x2/33/2. Applying the usual laws of exponents, we’d be tempted to simplify this as x2/33/2 3 · 3 2 = x1
= x. = x √ −12 However, if we substitute x = −1 and apply Deﬁnition 5.5, we ﬁnd (−1)2/3 = 3 = (−1)2 = 1 so that (−1)2/33/2 = x. If we take the time to rewrite x2/33/2 = 13 = 1. We see in this case that x2/33/2 = 13/2 = √ with radicals, we see 13 2 x2/33/2 √ x23/2 3 = = 3 √ 3 x2 √ = 3 3 x = √ 3 x3 = \|x\| 2Otherwise we’d run into the same paradox we did in Section 3.4. 5.3 Other Algebraic Functions 399 In the play-by-play analysis, we see that when we canceled the 2’s in multiplying 2 2, we were, x2 = \|x\| and not in fact, attempting to cancel a square with a square root. The fact that simply x is the root3 of the trouble. It may amuse the reader to know that x3/22/3 = x, and this veriﬁcation is left as an exercise. The moral of the story is that when simplifying fractional exponents, it’s usually best to rewrite them as radicals.4 The last major property we will state, and leave to Calculus to prove, is that radical functions are continuous on their domains, so the Intermediate Value Theorem, Theorem 3.1, applies. This means that if we take combinations of radical functions with polynomial and rational functions to form what the authors consider the algebraic functions,5 we can make sign diagrams using the procedure set forth in Section 4.2. 3 · 3 √ Steps for Constructing a Sign Diagram for an Algebraic Function Suppose f is an algebraic function. 1. Place any values excluded from the domain of f on the number line with an ‘’ above them. 2. Find the zeros of f and place them on the number line with the number 0 above them. 3. Choose a test value in each of the intervals determined in steps 1 and 2. 4. Determine the sign of f (x) for each test value in step 3, and write that sign above the corresponding interval. Our next example reviews quite a bit of Intermediate Algebra and demonstrates
some of the new features of these graphs. Example 5.3.1. For the following functions, state their domains and create sign diagrams. Check your answer graphically using your calculator. √ 1. f (x) = 3x 3 2 − x 3. h(x) = 3 8x x + 1 Solution. 2. g(x. k(x) = √ 2x x2 − 1 1. As far as domain is concerned, f (x) has no denominators and no even roots, which means its domain is (−∞, ∞). To create the sign diagram, we ﬁnd the zeros of f. 3Did you like that pun? 4In most other cases, though, rational exponents are preferred. 5As mentioned in Section 2.2, f (x) = √ x2 = \|x\| so that absolute value is also considered an algebraic function. 400 Further Topics in Functions √ 3x 3 f (x 3x = 0 or √ x = 0 or 3 2 − x3 x = 0 or 2 − x = 0 x = 0 or x = 2 = 03 The zeros 0 and 2 divide the real number line into three test intervals. The sign diagram and accompanying graph are below. Note that the intervals on which f is (+) correspond to where the graph of f is above the x-axis, and where the graph of f is below the x-axis we have that f is (−). The calculator suggests something mysterious happens near x = 2. Zooming in shows the graph becomes nearly vertical there. You’ll have to wait until Calculus to fully understand this phenomenon. (−) 0 (+) 0 (−) 0 2 y = f (x) y = f (x) near x = 2. 2. In g(x using a sign diagram. If we let r(x) = 2 − 4 x + 3, we have two radicals both of which are even indexed. To satisfy √ x + 3, we need 2 − 4 x + 3 ≥ 0. x + 3, we require x + 3 ≥ 0 or x ≥ −3. To satisfy √ While it may be tempting to write this as 2 ≥ 4 x + 3 and take both sides to the fourth power, there are times when this technique will produce erroneous results.6 Instead, we solve √ 2 − 4 x + 3, we know x ≥ −3, so we concern ourselves with
only this portion of the number line. To ﬁnd the zeros of r we set √ √ = 24 from which r(x) = 0 and solve 2 − 4 x + 3 = 0. We get 4 we obtain x + 3 = 16 or x = 13. Since we raised both sides of an equation to an even power, we need to check to see if x = 13 is an extraneous solution.7 We ﬁnd x = 13 does check since √ 2 − 4 16 = 2 − 2 = 0. Below is our sign diagram for r. √ x + 3 = 2 so that 4 √ 13 + + 34 (+) 0 (−) −3 13 √ We ﬁnd 2 − 4 we look for the zeros of g. Setting g(x) = 0 is equivalent to x + 3 ≥ 0 on [−3, 13] so this is the domain of g. To ﬁnd a sign diagram for g, x + 3 = 0. After squaring √ 2 − 4 √ 6For instance, −2 ≥ 4 7Recall, this means we have produced a candidate which doesn’t satisfy the original equation. Do you remember √ x + 3, which has no solution or −2 ≤ 4 x + 3 whose solution is [−3, ∞). how raising both sides of an equation to an even power could cause this? 5.3 Other Algebraic Functions 401 √ both sides, we get 2 − 4 x + 3 = 0, whose solution we have found to be x = 13. Since we squared both sides, we double check and ﬁnd g(13) is, in fact, 0. Our sign diagram and graph of g are below. Since the domain of g is [−3, 13], what we have below is not just a portion of the graph of g, but the complete graph. It is always above or on the x-axis, which veriﬁes our sign diagram. (+) −3 13 The complete graph of y = g(x). 8x x+1 = 0, we cube both sides to get 3. The radical in h(x) is odd, so our only concern is the denominator. Setting x + 1 = 0 gives x = −1, so our domain is (−∞, −1) ∪ (−1, ∞). To ﬁnd the zeros
of h, we set h(x) = 0. To 8x solve 3 x+1 = 0. We get 8x = 0, or x = 0. Below is the resulting sign diagram and corresponding graph. From the graph, it appears as though x = −1 is a vertical asymptote. Carrying out an analysis as x → −1 as in Section 4.2 conﬁrms this. (We leave the details to the reader.) Near x = 0, we have a situation similar to x = 2 in the graph of f in number 1 above. Finally, it appears as if the graph of h has a horizontal 8x asymptote y = 2. Using techniques from Section 4.2, we ﬁnd as x → ±∞, x+1 → 8. From this, it is hardly surprising that as x → ±∞, h(x) = 3 8x √ x+1 ≈ 3 8 = 2. (+) (−) 0 (+) −1 0 4. To ﬁnd the domain of k, we have both an even root and a denominator to concern ourselves with. To satisfy the square root, x2 − 1 ≥ 0. Setting r(x) = x2 − 1, we ﬁnd the zeros of r to be x = ±1, and we ﬁnd the sign diagram of r to be y = h(x) (+) 0 (−) 0 (+) −1 1 402 Further Topics in Functions √ We ﬁnd x2 − 1 ≥ 0 for (−∞, −1] ∪ [1, ∞). To keep the denominator of k(x) away from zero, x2 − 1 = 0. We leave it to the reader to verify the solutions are x = ±1, both of we set which must be excluded from the domain. Hence, the domain of k is (−∞, −1) ∪ (1, ∞). To build the sign diagram for k, we need the zeros of k. Setting k(x) = 0 results in = 0. We get 2x = 0 or x = 0. However, x = 0 isn’t in the domain of k, which means k has no zeros. We construct our sign diagram on the domain of k below alongside the graph of k. It appears that the graph of k has two vertical asympt
otes, one at x = −1 and one at x = 1. The gap in the graph between the asymptotes is because of the gap in the domain of k. Concerning end behavior, there appear to be two horizontal asymptotes, y = 2 and y = −2. To see why this is the case, we think of x → ±∞. The radicand of the denominator x2 − 1 ≈ x2, and as such, k(x) = 2x√ x2 = 2x x = 2. On the other hand, as x → −∞, \|x\| = −x, and as such k(x) ≈ 2x −x = −2. Finally, it appears as though the graph of k passes the Horizontal Line Test which means k is one to one and k−1 exists. Computing k−1 is left as an exercise. \|x\|. As x → ∞, we have \|x\| = x so k(x) ≈ 2x ≈ 2x√ 2x√ x2−1 x2−1 (−) −1 (+) 1 y = k(x) As the previous example illustrates, the graphs of general algebraic functions can have features we’ve seen before, like vertical and horizontal asymptotes, but they can occur in new and exciting ways. For example, k(x) = 2x√ had two distinct horizontal asymptotes. You’ll recall that rational functions could have at most one horizontal asymptote. Also some new characteristics like ‘unusual steepness’8 and cusps9 can appear in the graphs of arbitrary algebraic functions. Our next example ﬁrst demonstrates how we can use sign diagrams to solve nonlinear inequalities. (Don’t panic. The technique is very similar to the ones used in Chapters 2, 3 and 4.) We then check our answers graphically with a calculator and see some of the new graphical features of the functions in this extended family. x2−1 Example 5.3.2. Solve the following inequalities. Check your answers graphically with a calculator. 8The proper Calculus term for this is ‘vertical tangent’, but for now we’ll be okay calling it ‘unusual steepness’. 9See page 241 for the ﬁrst reference to this
feature. 5.3 Other Algebraic Functions 403 1. x2/3 < x4/3 − 6 2. 3(2 − x)1/3 ≤ x(2 − x)−2/3 Solution. 1. To solve x2/3 < x4/3 − 6, we get 0 on one side and attempt to solve x4/3 − x2/3 − 6 > 0. We set r(x) = x4/3 − x2/3 − 6 and note that since the denominators in the exponents are 3, they correspond to cube roots, which means the domain of r is (−∞, ∞). To ﬁnd the zeros for the sign diagram, we set r(x) = 0 and attempt to solve x4/3 − x2/3 − 6 = 0. At this point, it may be unclear how to proceed. We could always try as a last resort converting back to radical notation, but in this case we can take a cue from Example 3.3.4. Since there are three terms, and the exponent on one of the variable terms, x4/3, is exactly twice that of the other, x2/3, we have ourselves a ‘quadratic in disguise’ and we can rewrite x4/3 − x2/3 − 6 = 0 as x2/32 − x2/3 − 6 = 0. If we let u = x2/3, then in terms of u, we get u2 − u − 6 = 0. Solving for u, we obtain u = −2 or u = 3. Replacing x2/3 back in for u, we get x2/3 = −2 or x2/3 = 3. To avoid the trouble we encountered in the discussion following Deﬁnition 5.5, we now convert back to radical notation. By interpreting x2/3 as 3√ x2 = −2 or 3√ x2 = 3. Cubing both sides of these equations results in x2 = −8, which admits no √ real solution, or x2 = 27, which gives x = ±3 3. We construct a sign diagram and ﬁnd x4/3 − x2/3 − 6 > 0 on −∞, −3 3, ∞. To check our answer graphically, we set f (x
) = x2/3 and g(x) = x4/3 − 6. The solution to x2/3 < x4/3 − 6 corresponds to the inequality f (x) < g(x), which means we are looking for the x values for which the graph of f is below the graph of g. Using the ‘Intersect’ command we conﬁrm10 that the graphs cross at x = ±3 3. We see that the graph of f is below the graph of g (the thicker curve) on √ −∞, −3 x2 we have 3√ √ 3 ∪ 3 3 ∪ 3 3, ∞. √ √ √ (+) −3 0 (−) √ 3 0 (+) √ 3 3 y = f (x) and y = g(x) As a point of interest, if we take a closer look at the graphs of f and g near x = 0 with the axes oﬀ, we see that despite the fact they both involve cube roots, they exhibit diﬀerent behavior near x = 0. The graph of f has a sharp turn, or cusp, while g does not.11 10Or at least conﬁrm to several decimal places 11Again, we introduced this feature on page 241 as a feature which makes the graph of a function ‘not smooth’. 404 Further Topics in Functions y = f (x) near x = 0 y = g(x) near x = 0 2. To solve 3(2 − x)1/3 ≤ x(2 − x)−2/3, we gather all the nonzero terms on one side and obtain 3(2 − x)1/3 − x(2 − x)−2/3 ≤ 0. We set r(x) = 3(2 − x)1/3 − x(2 − x)−2/3. As in number 1, the denominators of the rational exponents are odd, which means there are no domain concerns there. However, the negative exponent on the second term indicates a denominator. Rewriting r(x) with positive exponents, we obtain r(x) = 3(2 − x)1/3 − x (2 − x)2/3 Setting the denominator equal to zero we get (2 − x)2/3 = 0, or 3 (2 − x
)2 = 0. After cubing both sides, and subsequently taking square roots, we get 2 − x = 0, or x = 2. Hence, the domain of r is (−∞, 2) ∪ (2, ∞). To ﬁnd the zeros of r, we set r(x) = 0. There are two school of thought on how to proceed and we demonstrate both. Factoring Approach. From r(x) = 3(2 − x)1/3 − x(2 − x)−2/3, we note that the quantity (2 − x) is common to both terms. When we factor out common factors, we factor out the quantity with the smaller exponent. In this case, since − 2 3, we factor (2 − x)−2/3 from both quantities. While it may seem odd to do so, we need to factor (2 − x)−2/3 from (2 − x)1/3, which results in subtracting the exponent − 2 3. We proceed using the usual properties of exponents.12 3 from 1 3 < 1 r(x) = 3(2 − x)1/3 − x(2 − x)−2/3 3 −(− 2 1 3(2 − x) = (2 − x)−2/3 = (2 − x)−2/3 3(2 − x)3/3 − x = (2 − x)−2/3 3(2 − x)1 − x = (2 − x)−2/3 (6 − 4x) = (2 − x)−2/3 (6 − 4x) 3 ) − x since 3√ √ u3 = ( 3 u)3 = u To solve r(x) = 0, we set (2 − x)−2/3 (6 − 4x) = 0, or or x = 3 2. 6−4x (2−x)2/3 = 0. We have 6 − 4x = 0 12And we exercise special care when reducing the 3 3 power to 1. 5.3 Other Algebraic Functions 405 Common Denominator Approach. We rewrite r(x) = 3(2 − x)1/3 − x(2 − x)−2/3 = 3(2 − x)1/3 − x (2 − x)2/3 x (
2 − x)2/3 1 3 − − = = = 3(2 − x)1/3(2 − x)2/3 (2 − x)2/3 3 + 2 3(2 − x) (2 − x)2/3 3(2 − x)3/3 (2 − x)2/3 3(2 − x)1 (2 − x)2/3 3(2 − x) − x (2 − x)2/3 6 − 4x (2 − x)2/3 As before, when we set r(x) = 0 we obtain x = 3 2. x (2 − x)2/3 x (2 − x)2/3 x (2 − x)2/3 = = = − − common denominator since 3√ √ u3 = ( 3 u)3 = u We now create our sign diagram and ﬁnd 3(2 − x)1/3 − x(2 − x)−2/3 ≤ 0 on 3 2, 2 ∪ (2, ∞). To check this graphically, we set f (x) = 3(2 − x)1/3 and g(x) = x(2 − x)−2/3 (the thicker curve). We conﬁrm that the graphs intersect at x = 3 2 and the graph of f is below the graph of g for x ≥ 3 2, with the exception of x = 2 where it appears the graph of g has a vertical asymptote. (+) 0 (−) (−) 3 2 2 y = f (x) and y = g(x) One application of algebraic functions was given in Example 1.6.6 in Section 1.1. Our last example is a more sophisticated application of distance. Example 5.3.3. Carl wishes to get high speed internet service installed in his remote Sasquatch observation post located 30 miles from Route 117. The nearest junction box is located 50 miles downroad from the post, as indicated in the diagram below. Suppose it costs $15 per mile to run cable along the road and $20 per mile to run cable oﬀ of the road. 406 Further Topics in Functions Outpost Route 117 50 miles Junction Box 1. Express the total cost C of connecting the Junction Box to the Outpost as a function of x, the number of miles the cable is run along Route 117
before heading oﬀ road directly towards the Outpost. Determine a reasonable applied domain for the problem. 2. Use your calculator to graph y = C(x) on its domain. What is the minimum cost? How far along Route 117 should the cable be run before turning oﬀ of the road? Solution. 1. The cost is broken into two parts: the cost to run cable along Route 117 at $15 per mile, and the cost to run it oﬀ road at $20 per mile. Since x represents the miles of cable run along Route 117, the cost for that portion is 15x. From the diagram, we see that the number of miles the cable is run oﬀ road is z, so the cost of that portion is 20z. Hence, the total cost is C = 15x + 20z. Our next goal is to determine z as a function of x. The diagram suggests we can use the Pythagorean Theorem to get y2 + 302 = z2. But we also see x + y = 50 so that y = 50 − x. Hence, z2 = (50 − x)2 + 900. Solving for z, we obtain z = ± (50 − x)2 + 900. (50 − x)2 + 900 so that our cost as a function Since z represents a distance, we choose z = of x only is given by C(x) = 15x + 20 (50 − x)2 + 900 From the context of the problem, we have 0 ≤ x ≤ 50. 2. Graphing y = C(x) on a calculator and using the ‘Minimum’ feature, we ﬁnd the relative minimum (which is also the absolute minimum in this case) to two decimal places to be (15.98, 1146.86). Here the x-coordinate tells us that in order to minimize cost, we should run 15.98 miles of cable along Route 117 and then turn oﬀ of the road and head towards the outpost. The y-coordinate tells us that the minimum cost, in dollars, to do so is $1146.86. The ability to stream live SasquatchCasts? Priceless. 5.3 Other Algebraic Functions 407 5.3.1 Exercises For each function in Exercises 1 - 10 below Find its domain. Create a sign diagram. Use your calculator to help you sketch its
graph and identify any vertical or horizontal asymp- totes, ‘unusual steepness’ or cusps. 1. f (x) = √ 1 − x2 √ 3. f (x) = x 1 − x2 5. f (x) = 4 16x x2 − 9 7. f (x) = x 2 3 (x − 7) 1 3 9. f (x) = x(x + 5)(x − 4) 2. f (x) = 4. f (x) = x √ x2 − 1 √ x2 − 1 6. f (x) = 3√ 5x x3 + 8 3 2 (x − 7) 1 3 8. f (x) = x 10. f (x) = 3√ x3 + 3x2 − 6x − 8 In Exercises 11 - 16, sketch the graph of y = g(x) by starting with the graph of y = f (x) and using the transformations presented in Section 1.7. √ 11. f (x) = 3 √ 13. f (x) = 4 √ 15. f (x) = 5 √ x, g(x) = 3 √ x, g(x) = 4 √ x, g(x √ 12. f (x) = 3 √ 14. f (x) = 4 √ 16. f (x) = 8 In Exercises 17 - 35, solve the equation or inequality. √ x, g(x) = −2 3 √ x, g(x) = 3 4 √ x, g(x) = 8 − 17. x + 1 = √ 3x + 7 √ 19. x + 3x + 10 = −2 21. 2x − 23. x √ 25. 2x + 1 = 3 + √ 4 − x √ 27. 10 − x − 2 ≤ 11 18. 2x + 1 = √ 3 − 3x 20. 3x + √ 6 − 9x = 2 22. x 3 2 = 8 √ 24 26. 5 − (4 − 2x 28. 408 Further Topics in Functions 29. 2(x − 2)− 1 3 − 2 3 x(x − 2)− 4 3 ≤ 0 31. 2x− 1 3 (x − 3) 1 3 +
x 2 3 (x − 3)− 2 3 ≥ 0 30. − 4 3√ 32. 3 (x − 2)− 4 3 + 8 9 x(x − 2)− 7 3 ≥ 0 x3 + 3x2 − 6x − 8 > x + 1 3 4 (x − 3)− 2 4 (x − 3) 1 3 < 0 33. 1 3 x 34. x− 1 3 + 3 4 x− 1 3 − x− 4 3 (x − 3)− 2 3 (x − 3)− 5 3 (x2 − 3x + 2) ≥ 0 35. 2 3 (x + 4) 3 5 (x − 2)− 1 3 + 3 5 (x + 4)− 2 5 (x − 2) 2 3 ≥ 0 36. Rework Example 5.3.3 so that the outpost is 10 miles from Route 117 and the nearest junction box is 30 miles down the road for the post. 37. The volume V of a right cylindrical cone depends on the radius of its base r and its height h 3 πr2h. The surface area S of a right cylindrical cone also r2 + h2. Suppose a cone is to have a and is given by the formula V = 1 depends on r and h according to the formula S = πr volume of 100 cubic centimeters. √ (a) Use the formula for volume to ﬁnd the height h as a function of r. (b) Use the formula for surface area and your answer to 37a to ﬁnd the surface area S as a function of r. (c) Use your calculator to ﬁnd the values of r and h which minimize the surface area. What is the minimum surface area? Round your answers to two decimal places. 38. The National Weather Service uses the following formula to calculate the wind chill: W = 35.74 + 0.6215 Ta − 35.75 V 0.16 + 0.4275 Ta V 0.16 where W is the wind chill temperature in ◦F, Ta is the air temperature in ◦F, and V is the wind speed in miles per hour. Note that W is deﬁned only for air temperatures at or lower than 50◦F and wind speeds above 3 miles per hour. (a) Suppose the air temperature is 42◦ and the wind speed is 7 miles per hour. Find the
wind chill temperature. Round your answer to two decimal places. (b) Suppose the air temperature is 37◦F and the wind chill temperature is 30◦F. Find the wind speed. Round your answer to two decimal places. 39. As a follow-up to Exercise 38, suppose the air temperature is 28◦F. (a) Use the formula from Exercise 38 to ﬁnd an expression for the wind chill temperature as a function of the wind speed, W (V ). (b) Solve W (V ) = 0, round your answer to two decimal places, and interpret. (c) Graph the function W using your calculator and check your answer to part 39b. 5.3 Other Algebraic Functions 409 40. The period of a pendulum in seconds is given by T = 2π L g (for small displacements) where L is the length of the pendulum in meters and g = 9.8 meters per second per second is the acceleration due to gravity. My Seth-Thomas antique schoolhouse clock needs T = 1 2 second and I can adjust the length of the pendulum via a small dial on the bottom of the bob. At what length should I set the pendulum? 41. The Cobb-Douglas production model states that the yearly total dollar value of the production output P in an economy is a function of labor x (the total number of hours worked in a year) and capital y (the total dollar value of all of the stuﬀ purchased in order to make things). Speciﬁcally, P = axby1−b. By ﬁxing P, we create what’s known as an ‘isoquant’ and we can then solve for y as a function of x. Let’s assume that the Cobb-Douglas production model for the country of Sasquatchia is P = 1.23x0.4y0.6. (a) Let P = 300 and solve for y in terms of x. If x = 100, what is y? (b) Graph the isoquant 300 = 1.23x0.4y0.6. What information does an ordered pair (x, y) which makes P = 300 give you? With the help of your classmates, ﬁnd several diﬀerent combinations of labor and capital all of which yield P = 300. Discuss any patterns you may see. 42.
According to Einstein’s Theory of Special Relativity, the observed mass m of an object is a function of how fast the object is traveling. Speciﬁcally, m(x) = mr 1 − x2 c2 where m(0) = mr is the mass of the object at rest, x is the speed of the object and c is the speed of light. (a) Find the applied domain of the function. (b) Compute m(.1c), m(.5c), m(.9c) and m(.999c). (c) As x → c−, what happens to m(x)? (d) How slowly must the object be traveling so that the observed mass is no greater than 100 times its mass at rest? 43. Find the inverse of k(x) = √ 2x x2 − 1. 410 Further Topics in Functions 44. Suppose Fritzy the Fox, positioned at a point (x, y) in the ﬁrst quadrant, spots Chewbacca the Bunny at (0, 0). Chewbacca begins to run along a fence (the positive y-axis) towards his warren. Fritzy, of course, takes chase and constantly adjusts his direction so that he is always running directly at Chewbacca. If Chewbacca’s speed is v1 and Fritzy’s speed is v2, the path Fritzy will take to intercept Chewbacca, provided v2 is directly proportional to, but not equal to, v1 is modeled by 1 2 y = x1+v1/v2 1 + v1/v2 − x1−v1/v2 1 − v1/v2 + v1v2 2 − v2 v2 1 (a) Determine the path that Fritzy will take if he runs exactly twice as fast as Chewbacca; that is, v2 = 2v1. Use your calculator to graph this path for x ≥ 0. What is the signiﬁcance of the y-intercept of the graph? (b) Determine the path Fritzy will take if Chewbacca runs exactly twice as fast as he does; that is, v1 = 2v2. Use your calculator to graph this path for x > 0. Describe the behavior of y as x → 0+ and interpret this physically. (c
) With the help of your classmates, generalize parts (a) and (b) to two cases: v2 > v1 and v2 < v1. We will discuss the case of v1 = v2 in Exercise 32 in Section 6.5. 45. Verify the Quotient Rule for Radicals in Theorem 5.6. 46. Show that 3 2 x 2 3 = x for all x ≥ 0. √ 47. Show that 3 2 is an irrational number by ﬁrst showing that it is a zero of p(x) = x3 − 2 and then showing p has no rational zeros. (You’ll need the Rational Zeros Theorem, Theorem 3.9, in order to show this last part.) √ 48. With the help of your classmates, generalize Exercise 47 to show that n c is an irrational number for any natural numbers c ≥ 2 and n ≥ 2 provided that c = pn for some natural number p. 5.3 Other Algebraic Functions 411 5.3.2 Answers √ 1. f (x) = 1 − x2 Domain: [−1, 1] (+) 0 −1 0 1 No asymptotes Unusual steepness at x = −1 and x = 1 No cusps √ 2. f (x) = x2 − 1 Domain: (−∞, −1] ∪ [1, ∞) 0 (+) (+) 0 −1 1 No asymptotes Unusual steepness at x = −1 and x = 1 No cusps √ 3. f (x) = x 1 − x2 Domain: [−1, 1] (−) 0 0 (+) −1 0 0 1 No asymptotes Unusual steepness at x = −1 and x = 1 No cusps √ 4. f (x) = x x2 − 1 Domain: (−∞, −1] ∪ [1, ∞) 0 (−) (+) 0 −1 1 No asymptotes Unusual steepness at x = −1 and x = 1 No cusps y 1 −1 1 x y 3 2 1 −3 −2 −1 1 2 3 x y 1 −1 1 x −1 y 3 2 1 −3 −2 −1 1 2 3 x −1 −2 −3 412 Further Topics in Functions 5. f (x) = 4
16x x2 − 9 Domain: (−3, 0] ∪ (3, ∞) 0 (+) (+) −3 0 3 Vertical asymptotes: x = −3 and x = 3 Horizontal asymptote: y = 0 Unusual steepness at x = 0 No cusps 6. f (x) = 3√ 5x x3 + 8 Domain: (−∞, −2) ∪ (−2, ∞) (+) (−) 0 (+) −2 0 Vertical asymptote x = −2 Horizontal asymptote y = 5 No unusual steepness or cusps 7. f (x) = x 1 3 Domain: (−∞, ∞) 2 3 (x − 7) (−) 0 0 (−) 0 (+) 7 No vertical or horizontal asymptotes13 Unusual steepness at x = 7 Cusp at x = 0 8. f (x) = x 3 2 (x − 7) 1 3 Domain: [0, ∞) 0 0 (−) 0 (+) 7 No asymptotes Unusual steepness at x = 7 No cusps y 5 4 3 2 1 −3 −2 −4−3−2−1 −1 −2 −3 −4 −5 −6 y 5 4 3 2 1 −3−2−1 −2 −3 −4 y 25 20 15 10 5 −5 −10 −15 1 2 3 4 5 6 7 8 x 13Using Calculus it can be shown that y = x − 7 3 is a slant asymptote of this graph. 5.3 Other Algebraic Functions 413 9. f (x) = x(x + 5)(x − 4) Domain: [−5, 0] ∪ [4, ∞) 0 (+) 0 0 (+) −5 0 4 No asymptotes Unusual steepness at x = −5, x = 0 and x = 4 No cusps 10. f (x) = 3√ x3 + 3x2 − 6x − 8 Domain: (−∞, ∞) 0 (−) 0 (+) (−) 0 (+) −4 −1 2 No vertical or horizontal asymptotes14 Unusual steepness at x = −4, x = −1 and x = 2 No cusps √ 11. g(x9 −7 −5 −3 −
1 −1 1 3 5 7 9 11 x −2 −3 −5−4−3−2−5−4−3−2−1 −1 1 2 3 4 5 x −2 −3 −4 √ 12. g(x) = − √ 13. g(x) = 4 x − 1 − 2 √ 14. g(x5 −3 −1 1 3 5 7 x y −1 −2 1 3 5 7 9 11 13 15 17 19 21 x y 5 4 3 2 1 −1 7 8 23 x 14Using Calculus it can be shown that y = x + 1 is a slant asymptote of this graph. 414 Further Topics in Functions √ 15. g(x) = 5 x + 2 + 3 √ 16. g(x) = 8 −34 −2 x 30 17. x = 3 18. x = 1 4 20. x = − 1 3, 2 3 21. x = 5+ √ 8 23. x = ±8 26. x = −2, 6 29. (−∞, 2) ∪ (2, 3] 32. (−∞, −1) 24. x = 6 27. [2, ∞) 30. (2, 6] 33. 0, 27 13 35. (−∞, −4) ∪ −4, − 22 19 ∪ (2, ∞) −40 −30 −20 −10 y x −1 −2 57 19. x = −3 22. x = 4 25. x = 4 28. [−1, 0] ∪ [1, ∞) 31. (−∞, 0) ∪ [2, 3) ∪ (3, ∞) 34. (−∞, 0) ∪ (0, 3) 36. C(x) = 15x + 20 100 + (30 − x)2, 0 ≤ x ≤ 30. The calculator gives the absolute minimum at ≈ (18.66, 582.29). This means to minimize the cost, approximately 18.66 miles of cable should be run along Route 117 before turning oﬀ the road and heading towards the outpost. The minimum cost to run the cable is approximately $582.29. 37. (a) h(r) = 300 πr2, r > 0. (b) S(r) = πr r2 + 300
πr2 √ 2 = π2r6+90000 r, r > 0 (c) The calculator gives the absolute minimum at the point ≈ (4.07, 90.23). This means the radius should be (approximately) 4.07 centimeters and the height should be 5.76 centimeters to give a minimum surface area of 90.23 square centimeters. 38. (a) W ≈ 37.55◦F. (b) V ≈ 9.84 miles per hour. 39. (a) W (V ) = 53.142 − 23.78V 0.16. Since we are told in Exercise 38 that wind chill is only eﬀect for wind speeds of more than 3 miles per hour, we restrict the domain to V > 3. (b) W (V ) = 0 when V ≈ 152.29. This means, according to the model, for the wind chill temperature to be 0◦F, the wind speed needs to be 152.29 miles per hour. 5.3 Other Algebraic Functions 415 (c) The graph is below. 40. 9.8 2 1 4π ≈ 0.062 meters or 6.2 centimeters 41. (a) First rewrite the model as P = 1.23x 2 5 y 3 5. Then 300 = 1.23x 2 5 y 3 5 yields y = If x = 100 then y ≈ 441.93687. 300 1.23x 2 5 5 3. 42. (a) [0, c) (b) m(.1c) = m(.9c) = mr√.99 mr√.19 ≈ 1.005mr m(.5c) = ≈ 2.294mr m(.999c) = ≈ 22.366mr mr√.75 √ ≈ 1.155mr mr.0.001999 (c) As x → c−, m(x) → ∞ (d) If the object is traveling no faster than approximately 0.99995 times the speed of light, then its observed mass will be no greater than 100mr. 43. k−1(x) = √ x x2 − 4 √ 44. (a) y = 1 3 x3/2 − x + 2 3. The point 0, 2 3 is when Fritzy’s path crosses
Chewbacca’s path - in other words, where Fritzy catches Chewbacca. (b) y = 1 2x − 2 6 x3 + 1 3. Using the techniques from Chapter 4, we ﬁnd as x → 0+, y → ∞ which means, in this case, Fritzy’s pursuit never ends; he never catches Chewbacca. This makes sense since Chewbacca has a head start and is running faster than Fritzy. y = 1 3 x3/ x3 + 1 2x − 2 3 416 Further Topics in Functions Chapter 6 Exponential and Logarithmic Functions 6.1 Introduction to Exponential and Logarithmic Functions Of all of the functions we study in this text, exponential and logarithmic functions are possibly the ones which impact everyday life the most.1 This section introduces us to these functions while the rest of the chapter will more thoroughly explore their properties. Up to this point, we have dealt with functions which involve terms like x2 or x2/3, in other words, terms of the form xp where the base of the term, x, varies but the exponent of each term, p, remains constant. In this chapter, we study functions of the form f (x) = bx where the base b is a constant and the exponent x is the variable. We start our exploration of these functions with f (x) = 2x. (Apparently this is a tradition. Every College Algebra book we have ever read starts with f (x) = 2x.) We make a table of values, plot the points and connect the dots in a pleasing fashion. x −3 −2 −1 0 1 2 3 f (x) 2−3 = 1 8 2−2 = 1 4 2−1 = 1 2 20 = 1 21 = 2 22 = 4 23 = 8 (x, f (x)) −3, 1 8 −2, 1 4 −1, 1 2 (0, 1) (1, 2) (2, 4) (3, 83 −2 −1 1 2 3 x y = f (x) = 2x A few remarks about the graph of f (x) = 2x which we have constructed are in order. As x → −∞ 1Take a class in Diﬀerential Equations and you’ll see why. 418 Exponential and Logarithmic Functions and attains values like x
= −100 or x = −1000, the function f (x) = 2x takes on values like f (−100) = 2−100 = 1 2100 or f (−1000) = 2−1000 = 1 21000. In other words, as x → −∞, 2x ≈ 1 very big (+) ≈ very small (+) √ √ √ So as x → −∞, 2x → 0+. This is represented graphically using the x-axis (the line y = 0) as a horizontal asymptote. On the ﬂip side, as x → ∞, we ﬁnd f (100) = 2100, f (1000) = 21000, and so on, thus 2x → ∞. As a result, our graph suggests the range of f is (0, ∞). The graph of f passes the Horizontal Line Test which means f is one-to-one and hence invertible. We also note that when we ‘connected the dots in a pleasing fashion’, we have made the implicit assumption that f (x) = 2x is continuous2 and has a domain of all real numbers. In particular, we have suggested that things 3 might 3 exist as real numbers. We should take a moment to discuss what something like 2 like 2 mean, and refer the interested reader to a solid course in Calculus for a more rigorous explanation. 3 = 1.73205... is an irrational number3 and as such, its decimal representation The number 3 by terminating decimals, and neither repeats nor terminates. We can, however, approximate √ it stands to reason4 we can use these to approximate 2 3. For example, if we approximate 3 100 = 100√ 2173. It is not, by any means, a pleasant by 1.73, we can approximate 2 number, but it is at least a number that we understand in terms of powers and roots. It also stands √ to reason that better and better approximations of 3 yield better and better approximations of 2 Suppose we wish to study the family of functions f (x) = bx. Which bases b make sense to study? We ﬁnd that we run into diﬃculty if b < 0. For example, if b = −2, then the function f (x) = (−2)x has trouble, for instance, at x =
1 −2 is not a real number. In general, if x is any rational number with an even denominator, then (−2)x is not deﬁned, so we must restrict our attention to bases b ≥ 0. What about b = 0? The function f (x) = 0x is undeﬁned for x ≤ 0 because we cannot divide by 0 and 00 is an indeterminant form. For x > 0, 0x = 0 so the function f (x) = 0x is the same as the function f (x) = 0, x > 0. We know everything we can possibly know about this function, so we exclude it from our investigations. The only other base we exclude is b = 1, since the function f (x) = 1x = 1 is, once again, a function we have already studied. We are now ready for our deﬁnition of exponential functions. 3 should be the result of this sequence of approximations.5 2 since (−2)1/2 = 3, so the value of 2 3 ≈ 21.73 = 2 √ √ 173 √ √ √ √ Deﬁnition 6.1. A function of the form f (x) = bx where b is a ﬁxed real number, b > 0, b = 1 is called a base b exponential function. We leave it to the reader to verify6 that if b > 1, then the exponential function f (x) = bx will share x the same basic shape and characteristics as f (x) = 2x. What if 0 < b < 1? Consider g(x) = 1. 2 We could certainly build a table of values and connect the points, or we could take a step back and 2Recall that this means there are no holes or other kinds of breaks in the graph. 3You can actually prove this by considering the polynomial p(x) = x2 − 3 and showing it has no rational zeros by applying Theorem 3.9. 4This is where Calculus and continuity come into play. 5Want more information? Look up “convergent sequences” on the Internet. 6Meaning, graph some more examples on your own. 6.1 Introduction to Exponential and Logarithmic Functions 419 note that g(x) = 1 2 the graph of f (−x) is obtained from the graph
of f (x) by reﬂecting it across the y-axis. We get = 2−x = f (−x), where f (x) = 2x. Thinking back to Section 1.7, = 2−1x 3−2−1 1 2 3 x y = f (x) = 2x reﬂect across y-axis −−−−−−−−−−−−→ multiply each x-coordinate by −1 −3−2−1 1 2 3 y = g(x) = 2−x = 1 2 x x We see that the domain and range of g match that of f, namely (−∞, ∞) and (0, ∞), respectively. Like f, g is also one-to-one. Whereas f is always increasing, g is always decreasing. As a result, as x → −∞, g(x) → ∞, and on the ﬂip side, as x → ∞, g(x) → 0+. It shouldn’t be too surprising that for all choices of the base 0 < b < 1, the graph of y = bx behaves similarly to the graph of g. We summarize the basic properties of exponential functions in the following theorem.7 Theorem 6.1. Properties of Exponential Functions: Suppose f (x) = bx. The domain of f is (−∞, ∞) and the range of f is (0, ∞). (0, 1) is on the graph of f and y = 0 is a horizontal asymptote to the graph of f. f is one-to-one, continuous and smootha If b > 1: If 0 < b < 1: – f is always increasing – As x → −∞, f (x) → 0+ – As x → ∞, f (x) → ∞ – The graph of f resembles: – f is always decreasing – As x → −∞, f (x) → ∞ – As x → ∞, f (x) → 0+ – The graph of f resembles: y = bx, b > 1 y = bx, 0 < b < 1 aRecall that this means the graph of f has no sharp turns or corners. 7The proof of which, like many things discussed in the text, requires Calculus. 420 Exponential and Logarithmic Functions
Of all of the bases for exponential functions, two occur the most often in scientiﬁc circles. The ﬁrst, base 10, is often called the common base. The second base is an irrational number, e ≈ 2.718, called the natural base. We will more formally discuss the origins of this number in Section 6.5. For now, it is enough to know that since e > 1, f (x) = ex is an increasing exponential function. The following examples give us an idea how these functions are used in the wild. Example 6.1.1. The value of a car can be modeled by V (x) = 25 4 5 car in years and V (x) is the value in thousands of dollars., where x ≥ 0 is age of the x 1. Find and interpret V (0). 2. Sketch the graph of y = V (x) using transformations. 3. Find and interpret the horizontal asymptote of the graph you found in 2. Solution. 1. To ﬁnd V (0), we replace x with 0 to obtain V (0) = 25 4 5 = 25. Since x represents the age of the car in years, x = 0 corresponds to the car being brand new. Since V (x) is measured in thousands of dollars, V (0) = 25 corresponds to a value of $25,000. Putting it all together, we interpret V (0) = 25 to mean the purchase price of the car was $25,000. 0 x 2. To graph y = 25 4 5, we start with the basic exponential function f (x) = 4. Since the 5 base b = 4 5 is between 0 and 1, the graph of y = f (x) is decreasing. We plot the y-intercept, and label the horizontal asymptote y = 0. (0, 1) and two other points, −1, 5 4 To obtain V (x) = 25 4, x ≥ 0, we multiply the output from f by 25, in other words, 5 V (x) = 25f (x). In accordance with Theorem 1.5, this results in a vertical stretch by a factor of 25. We multiply all of the y values in the graph by 25 (including the y value of the horizontal asymptote) and obtain the points −1, 125, (0, 25) and (1, 20). The
horizontal 4 asymptote remains y = 0. Finally, we restrict the domain to [0, ∞) to ﬁt with the applied domain given to us. We have the result below. and 1, 4 5 x x y 2 (0, 1) −3−2−1 1 2 3 x H.A. y = 0 y = f (x) = 4 5 x y 30 (0, 25) 20 15 10 5 vertical scale by a factor of 25 −−−−−−−−−−−−−−−−−−−−−→ multiply each y-coordinate by 25 1 2 3 4 5 6 x H.A. y = 0 y = V (x) = 25f (x), x ≥ 0 3. We see from the graph of V that its horizontal asymptote is y = 0. (We leave it to reader to verify this analytically by thinking about what happens as we take larger and larger powers of 4 5.) This means as the car gets older, its value diminishes to 0. 6.1 Introduction to Exponential and Logarithmic Functions 421 The function in the previous example is often called a ‘decay curve’. Increasing exponential functions are used to model ‘growth curves’ and we shall see several diﬀerent examples of those in Section 6.5. For now, we present another common decay curve which will serve as the basis for further study of exponential functions. Although it may look more complicated than the previous example, it is actually just a basic exponential function which has been modiﬁed by a few transformations from Section 1.7. Example 6.1.2. According to Newton’s Law of Cooling8 the temperature of coﬀee T (in degrees Fahrenheit) t minutes after it is served can be modeled by T (t) = 70 + 90e−0.1t. 1. Find and interpret T (0). 2. Sketch the graph of y = T (t) using transformations. 3. Find and interpret the horizontal asymptote of the graph. Solution. 1. To ﬁnd T (0), we replace every occurrence of the independent variable t with 0 to obtain T (0) = 70 + 90e−0.1(0) = 160. This means that the coﬀee was served at 160◦F. 2. To graph
y = T (t) using transformations, we start with the basic function, f (t) = et. As we have already remarked, e ≈ 2.718 > 1 so the graph of f is an increasing exponential with y-intercept (0, 1) and horizontal asymptote y = 0. The points −1, e−1 ≈ (−1, 0.37) and (1, e) ≈ (1, 2.72) are also on the graph. Since the formula T (t) looks rather complicated, we rewrite T (t) in the form presented in Theorem 1.7 and use that result to track the changes to our three points and the horizontal asymptote. We have T (t) = 70 + 90e−0.1t = 90e−0.1t + 70 = 90f (−0.1t) + 70 Multiplication of the input to f, t, by −0.1 results in a horizontal expansion by a factor of 10 as well as a reﬂection about the y-axis. We divide each of the x values of our points by −0.1 (which amounts to multiplying them by −10) to obtain 10, e−1, (0, 1), and (−10, e). Since none of these changes aﬀected the y values, the horizontal asymptote remains y = 0. Next, we see that the output from f is being multiplied by 90. This results in a vertical stretch by a factor of 90. We multiply the y-coordinates by 90 to obtain 10, 90e−1, (0, 90), and (−10, 90e). We also multiply the y value of the horizontal asymptote y = 0 by 90, and it remains y = 0. Finally, we add 70 to all of the y-coordinates, which shifts the graph upwards to obtain 10, 90e−1 + 70 ≈ (10, 103.11), (0, 160), and (−10, 90e + 70) ≈ (−10, 314.64). Adding 70 to the horizontal asymptote shifts it upwards as well to y = 70. We connect these three points using the same shape in the same direction as in the graph of f and, last but not least, we restrict the domain to match the applied domain [0, ∞). The result is below. 8We will discuss
this in greater detail in Section 6.5. 422 Exponential and Logarithmic Functions y 7 6 5 4 3 2 (0, 1) y 180 160 140 120 100 80 60 40 20 H.A. y = 70 −3−2−1 1 2 3 t H.A. y = 0 y = f (t) = et −−−−−−−−−−−−→ y = T (t) 2 4 6 8 10 12 14 16 18 20 t 3. From the graph, we see that the horizontal asymptote is y = 70. It is worth a moment or two of our time to see how this happens analytically and to review some of the ‘number sense’ developed in Chapter 4. As t → ∞, We get T (t) = 70 + 90e−0.1t ≈ 70 + 90every big (−). Since e > 1, every big (−) = 1 every big (+) ≈ 1 very big (+) ≈ very small (+) The larger t becomes, the smaller e−0.1t becomes, so the term 90e−0.1t ≈ very small (+). Hence, T (t) ≈ 70 + very small (+) which means the graph is approaching the horizontal line y = 70 from above. This means that as time goes by, the temperature of the coﬀee is cooling to 70◦F, presumably room temperature. As we have already remarked, the graphs of f (x) = bx all pass the Horizontal Line Test. Thus the exponential functions are invertible. We now turn our attention to these inverses, the logarithmic functions, which are called ‘logs’ for short. Deﬁnition 6.2. The inverse of the exponential function f (x) = bx is called the base b logarithm function, and is denoted f −1(x) = logb(x) We read ‘logb(x)’ as ‘log base b of x.’ We have special notations for the common base, b = 10, and the natural base, b = e. Deﬁnition 6.3. The common logarithm of a real number x is log10(x) and is usually written log(x). The natural logarithm of a real number x
is loge(x) and is usually written ln(x). Since logs are deﬁned as the inverses of exponential functions, we can use Theorems 5.2 and 5.3 to tell us about logarithmic functions. For example, we know that the domain of a log function is the range of an exponential function, namely (0, ∞), and that the range of a log function is the domain of an exponential function, namely (−∞, ∞). Since we know the basic shapes of y = f (x) = bx for the diﬀerent cases of b, we can obtain the graph of y = f −1(x) = logb(x) by reﬂecting the graph of f across the line y = x as shown below. The y-intercept (0, 1) on the graph of f corresponds to an x-intercept of (1, 0) on the graph of f −1. The horizontal asymptotes y = 0 on the graphs of the exponential functions become vertical asymptotes x = 0 on the log graphs. 6.1 Introduction to Exponential and Logarithmic Functions 423 y = bx, b > 1 y = logb(x), b > 1 y = bx, 0 < b < 1 y = logb(x), 0 < b < 1 On a procedural level, logs undo the exponentials. Consider the function f (x) = 2x. When we evaluate f (3) = 23 = 8, the input 3 becomes the exponent on the base 2 to produce the real number 8. The function f −1(x) = log2(x) then takes the number 8 as its input and returns the exponent 3 as its output. In symbols, log2(8) = 3. More generally, log2(x) is the exponent you put on 2 to get x. Thus, log2(16) = 4, because 24 = 16. The following theorem summarizes the basic properties of logarithmic functions, all of which come from the fact that they are inverses of exponential functions. Theorem 6.2. Properties of Logarithmic Functions: Suppose f (x) = logb(x). The domain of f is (0, ∞) and the range of f is (−∞, ∞). (1, 0) is on the graph
of f and x = 0 is a vertical asymptote of the graph of f. f is one-to-one, continuous and smooth ba = c if and only if logb(c) = a. That is, logb(c) is the exponent you put on b to obtain c. logb (bx) = x for all x and blogb(x) = x for all x > 0 If b > 1: If 0 < b < 1: – f is always increasing – As x → 0+, f (x) → −∞ – As x → ∞, f (x) → ∞ – f is always decreasing – As x → 0+, f (x) → ∞ – As x → ∞, f (x) → −∞ – The graph of f resembles: – The graph of f resembles: y = logb(x), b > 1 y = logb(x), 0 < b < 1 424 Exponential and Logarithmic Functions As we have mentioned, Theorem 6.2 is a consequence of Theorems 5.2 and 5.3. However, it is worth the reader’s time to understand Theorem 6.2 from an exponential perspective. For instance, we know that the domain of g(x) = log2(x) is (0, ∞). Why? Because the range of f (x) = 2x is (0, ∞). In a way, this says everything, but at the same time, it doesn’t. For example, if we try to ﬁnd log2(−1), we are trying to ﬁnd the exponent we put on 2 to give us −1. In other words, we are looking for x that satisﬁes 2x = −1. There is no such real number, since all powers of 2 are positive. While what we have said is exactly the same thing as saying ‘the domain of g(x) = log2(x) is (0, ∞) because the range of f (x) = 2x is (0, ∞)’, we feel it is in a student’s best interest to understand the statements in Theorem 6.2 at this level instead of just merely memorizing the facts. Example 6.1.3. Simplify the following. 1. log3(81) 5. log(
0.001) Solution. 2. log2 1 8 6. 2log2(8) 3. log√ 5(25) 7. 117− log117(6) 3√ e2 4. ln 1. The number log3(81) is the exponent we put on 3 to get 81. As such, we want to write 81 as a power of 3. We ﬁnd 81 = 34, so that log3(81) = 4., we need rewrite 1 2. To ﬁnd log2 8 as a power of 2. We ﬁnd 1 1 8 8 = 1 √ 23 = 2−3, so log2 5. We know 25 = 52, and = −3. 1 8 3. To determine log√ 52 5 = √, so we have 25 = √ 5(25), we need to express 25 as a power of = √. We get log√ e2 522 3√ e2 e2. Rewriting 3√ means loge e2 = e2/3, we ﬁnd ln 54 3√ 5(25) = 4. 4. First, recall that the notation ln to put on e to obtain 3√, so we are looking for the exponent = ln e2/3 = 2 3. e2 3√ 1000 = 1 have 0.001 = 1 5. Rewriting log(0.001) as log10(0.001), we see that we need to write 0.001 as a power of 10. We 103 = 10−3. Hence, log(0.001) = log 10−3 = −3. 6. We can use Theorem 6.2 directly to simplify 2log2(8) = 8. We can also understand this problem by ﬁrst ﬁnding log2(8). By deﬁnition, log2(8) is the exponent we put on 2 to get 8. Since 8 = 23, we have log2(8) = 3. We now substitute to ﬁnd 2log2(8) = 23 = 8. 7. From Theorem 6.2, we know 117log117(6) = 6, but we cannot directly apply this formula to the expression 117− log117(6). (Can you see why?) At this point, we use a property
of exponents followed by Theorem 6.2 to get9 117− log117(6) = 1 117log117(6) = 1 6 9It is worth a moment of your time to think your way through why 117log117(6) = 6. By deﬁnition, log117(6) is the exponent we put on 117 to get 6. What are we doing with this exponent? We are putting it on 117. By deﬁnition we get 6. In other words, the exponential function f (x) = 117x undoes the logarithmic function g(x) = log117(x). 6.1 Introduction to Exponential and Logarithmic Functions 425 Up until this point, restrictions on the domains of functions came from avoiding division by zero and keeping negative numbers from beneath even radicals. With the introduction of logs, we now have another restriction. Since the domain of f (x) = logb(x) is (0, ∞), the argument10 of the log must be strictly positive. Example 6.1.4. Find the domain of the following functions. Check your answers graphically using the calculator. 1. f (x) = 2 log(3 − x) − 1 Solution. 2. g(x) = ln x x − 1 1. We set 3−x > 0 to obtain x < 3, or (−∞, 3). The graph from the calculator below veriﬁes this. Note that we could have graphed f using transformations. Taking a cue from Theorem 1.7, we rewrite f (x) = 2 log10(−x + 3) − 1 and ﬁnd the main function involved is y = h(x) = log10(x). We select three points to track, 1 10, −1, (1, 0) and (10, 1), along with the vertical asymptote x = 0. Since f (x) = 2h(−x + 3) − 1, Theorem 1.7 tells us that to obtain the destinations of these points, we ﬁrst subtract 3 from the x-coordinates (shifting the graph left 3 units), then divide (multiply) by the x-coordinates by −1 (causing a reﬂection across the y-axis). These transformations apply to the vertical asymptote x = 0 as well
. Subtracting 3 gives us x = −3 as our asymptote, then multplying by −1 gives us the vertical asymptote x = 3. Next, we multiply the y-coordinates by 2 which results in a vertical stretch by a factor of 2, then we ﬁnish by subtracting 1 from the y-coordinates which shifts the graph down 1 unit. We leave it to the reader to perform the indicated arithmetic on the points themselves and to verify the graph produced by the calculator below. 2. To ﬁnd the domain of g, we need to solve the inequality x a sign diagram. If we deﬁne r(x) = x x = 0. Choosing some test values, we generate the sign diagram below. x−1 > 0. As usual, we proceed using x−1, we ﬁnd r is undeﬁned at x = 1 and r(x) = 0 when (+) 0 (−) (+) 0 1 We ﬁnd x x−1 > 0 on (−∞, 0) ∪ (1, ∞) to get the domain of g. The graph of y = g(x) conﬁrms this. We can tell from the graph of g that it is not the result of Section 1.7 transformations being applied to the graph y = ln(x), so barring a more detailed analysis using Calculus, the calculator graph is the best we can do. One thing worthy of note, however, is the end behavior of g. The graph suggests that as x → ±∞, g(x) → 0. We can verify this analytically. Using x x−1 ≈ 1. Hence, it makes results from Chapter 4 and continuity, we know that as x → ±∞, sense that g(x) = ln ≈ ln(1) = 0. x x−1 10See page 55 if you’ve forgotten what this term means. 426 Exponential and Logarithmic Functions y = f (x) = 2 log(3 − x) − 1 y = g(x) = ln x x − 1 While logarithms have some interesting applications of their own which you’ll explore in the exercises, their primary use to us will be to undo exponential functions. (This is, after all, how they were deﬁ
ned.) Our last example solidiﬁes this and reviews all of the material in the section. Example 6.1.5. Let f (x) = 2x−1 − 3. 1. Graph f using transformations and state the domain and range of f. 2. Explain why f is invertible and ﬁnd a formula for f −1(x). 3. Graph f −1 using transformations and state the domain and range of f −1. 4. Verify f −1 ◦ f (x) = x for all x in the domain of f and f ◦ f −1 (x) = x for all x in the domain of f −1. 5. Graph f and f −1 on the same set of axes and check the symmetry about the line y = x. Solution. 1. If we identify g(x) = 2x, we see f (x) = g(x − 1) − 3. We pick the points −1, 1 2, (0, 1) and (1, 2) on the graph of g along with the horizontal asymptote y = 0 to track through the transformations. By Theorem 1.7 we ﬁrst add 1 to the x-coordinates of the points on the graph of g (shifting g to the right 1 unit) to get 0, 1, (1, 1) and (2, 2). The horizontal 2 asymptote remains y = 0. Next, we subtract 3 from the y-coordinates, shifting the graph down 3 units. We get the points 0, − 5, (1, −2) and (2, −1) with the horizontal asymptote 2 now at y = −3. Connecting the dots in the order and manner as they were on the graph of g, we get the graph below. We see that the domain of f is the same as g, namely (−∞, ∞), but that the range of f is (−3, ∞). 3−2−1 −1 −2 −3 −3−2−1 −1 −2 1 2 3 4 x y = h(x) = 2x −−−−−−−−−−−−→ y = f (x) = 2x−1 − 3 6.1 Introduction to Exponential and Logarithmic Functions 427 2. The graph of f passes the Horizontal Line
Test so f is one-to-one, hence invertible. To ﬁnd a formula for f −1(x), we normally set y = f (x), interchange the x and y, then proceed to solve for y. Doing so in this situation leads us to the equation x = 2y−1 − 3. We have yet to discuss how to solve this kind of equation, so we will attempt to ﬁnd the formula for f −1 from a procedural perspective. If we break f (x) = 2x−1 − 3 into a series of steps, we ﬁnd f takes an input x and applies the steps (a) subtract 1 (b) put as an exponent on 2 (c) subtract 3 Clearly, to undo subtracting 1, we will add 1, and similarly we undo subtracting 3 by adding 3. How do we undo the second step? The answer is we use the logarithm. By deﬁnition, log2(x) undoes exponentiation by 2. Hence, f −1 should (a) add 3 (b) take the logarithm base 2 (c) add 1 In symbols, f −1(x) = log2(x + 3) + 1. 3. To graph f −1(x) = log2(x + 3) + 1 using transformations, we start with j(x) = log2(x). We 2, −1, (1, 0) and (2, 1) on the graph of j along with the vertical asymptote track the points 1 x = 0 through the transformations using Theorem 1.7. Since f −1(x) = j(x + 3) + 1, we ﬁrst subtract 3 from each of the x values (including the vertical asymptote) to obtain − 5 2, −1, (−2, 0) and (−1, 1) with a vertical asymptote x = −3. Next, we add 1 to the y values on the 2, 0, (−2, 1) and (−1, 2). If you are experiencing d´ej`a vu, there is a good graph and get − 5 reason for it but we leave it to the reader to determine the source of this uncanny familiarity. We obtain the graph below. The domain of f −1 is (−3, ∞), which matches the range
of f, and the range of f −1 is (−∞, ∞), which matches the domain of f. y 4 3 2 1 −3−2−1 −1 −2 −2−1 −1 −2 −(x) = log2(x) −−−−−−−−−−−−→ y = f −1(x) = log2(x + 3) + 1 4. We now verify that f (x) = 2x−1 − 3 and f −1(x) = log2(x + 3) + 1 satisfy the composition requirement for inverses. For all real numbers x, 428 Exponential and Logarithmic Functions f −1 ◦ f (x) = f −1(f (x)) 2x−1 − 3 + 3 + 1 2x−1 + 1 = f −1 2x−1 − 3 = log2 = log2 = (x − 1) + 1 = x For all real numbers x > −3, we have11 f ◦ f −1 (x) = f f −1(x) Since log2 (2u) = u for all real numbers u = f (log2(x + 3) + 1) = 2(log2(x+3)+1)−1 − 3 = 2log2(x+3) − 3 = (x + 3) − 3 = x Since 2log2(u) = u for all real numbers u > 0 5. Last, but certainly not least, we graph y = f (x) and y = f −1(x) on the same set of axes and see the symmetry about the line y = x3 −2 −1 −1 −2 y = f (x) = 2x−1 − 3 y = f −1(x) = log2(x + 3) + 1 11Pay attention - can you spot in which step below we need x > −3? 6.1 Introduction to Exponential and Logarithmic Functions 429 6.1.1 Exercises In Exercises 1 - 15, use the property: ba = c if and only if logb(c) = a from Theorem 6.2 to rewrite the given equation in the other form. That is, rewrite the exponential equations as logarithmic equations and rewrite the logarithmic equations as exponential equations. 1. 23 = 8 4
. 1 3 −2 = 9 7. e0 = 1 10. log3 1 81 = −4 2. 5−3 = 1 125 5. 4 25 −1/2 = 5 2 8. log5(25) = 2 11. log 4 3 3 4 = −1 13. log(0.1) = −1 14. ln(e) = 1 In Exercises 16 - 42, evaluate the expression. 16. log3(27) 19. log6 1 36 22. log 1 5 (625) 25. log 1 1000000 28. log4(8) 31. log36 √ 4 36 34. log36 36216 37. log 3√ 105 40. log eln(100) 17. log6(216) 20. log8(4) 23. log 1 6 (216) 26. log(0.01) 29. log6(1) 32. 7log7(3) 35. ln e5 38. ln 1√ e 41. log2 3− log3(2) 3. 45/2 = 32 6. 10−3 = 0.001 9. log25(5) = 1 2 12. log(100) = 2 15. ln 1√ e = − 1 2 18. log2(32) 21. log36(216) 24. log36(36) 27. ln e3 30. log13 √ 13 33. 36log36(216) 1011 36. log 9√ 39. log5 3log3(5) 42. ln 426 log(1) In Exercises 43 - 57, ﬁnd the domain of the function. 43. f (x) = ln(x2 + 1) 45. f (x) = ln(4x − 20) 44. f (x) = log7(4x + 8) 46. f (x) = log x2 + 9x + 18 430 Exponential and Logarithmic Functions 47. f (x) = log x + 2 x2 − 1 49. f (x) = ln(7 − x) + ln(x − 4) 51. f (x) = log x2 + x + 1 53. f (x) = log9(\|x + 3\| − 4) 55. f (x) = 1 3 − log5(x)
57. f (x) = ln(−2x3 − x2 + 13x − 6) 48. f (x) = log x2 + 9x + 18 4x − 20 50. f (x) = ln(4x − 20) + ln x2 + 9x + 18 52. f (x) = 4 54. f (x) = ln( √ 56. f (x) = log4(x) √ x − 4 − 3) −1 − x (x) log 1 2 In Exercises 58 - 63, sketch the graph of y = g(x) by starting with the graph of y = f (x) and using transformations. Track at least three points of your choice and the horizontal asymptote through the transformations. State the domain and range of g. 58. f (x) = 2x, g(x) = 2x − 1 60. f (x) = 3x, g(x) = 3−x + 2 59. f (x) = 1 3 x, g(x) = 1 3 x−1 61. f (x) = 10x, g(x) = 10 x+1 2 − 20 62. f (x) = ex, g(x) = 8 − e−x 63. f (x) = ex, g(x) = 10e−0.1x In Exercises 64 - 69, sketch the graph of y = g(x) by starting with the graph of y = f (x) and using transformations. Track at least three points of your choice and the vertical asymptote through the transformations. State the domain and range of g. 64. f (x) = log2(x), g(x) = log2(x + 1) 65. f (x) = log 1 3 (x), g(x) = log 1 3 (x) + 1 66. f (x) = log3(x), g(x) = − log3(x − 2) 67. f (x) = log(x), g(x) = 2 log(x + 20) − 1 68. f (x) = ln(x), g(x) = − ln(8 − x) 69. f (x) = ln(x), g(x) = −10 ln x 10 70.

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