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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 − 2x 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 dOpis 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 deﬁne 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 asymptotes, 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 con�
�rm10 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 �
�nd 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 Chewb
acca. 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 deﬁned.) 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 Exerc
ises 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. Verify that each function in Exercises 64 - 69 is the inverse of the corresponding function in Exercises 58 - 63. (Match up #58 and #64, and so on.) In Exercises 71 - 74, ﬁnd the inverse of the function from the ‘procedural perspective’ discussed in Example 6.1.5 and graph the function and its inverse on
the same set of axes. 71. f (x) = 3x+2 − 4 73. f (x) = −2−x + 1 72. f (x) = log4(x − 1) 74. f (x) = 5 log(x) − 2 6.1 Introduction to Exponential and Logarithmic Functions 431 (Logarithmic Scales) In Exercises 75 - 77, we introduce three widely used measurement scales which involve common logarithms: the Richter scale, the decibel scale and the pH scale. The computations involved in all three scales are nearly identical so pay attention to the subtle diﬀerences. 75. Earthquakes are complicated events and it is not our intent to provide a complete discussion of the science involved in them. Instead, we refer the interested reader to a solid course in Geology12 or the U.S. Geological Survey’s Earthquake Hazards Program found here and present only a simpliﬁed version of the Richter scale. The Richter scale measures the magnitude of an earthquake by comparing the amplitude of the seismic waves of the given earthquake to those of a “magnitude 0 event”, which was chosen to be a seismograph reading of 0.001 millimeters recorded on a seismometer 100 kilometers from the earthquake’s epicenter. Speciﬁcally, the magnitude of an earthquake is given by M (x) = log x 0.001 where x is the seismograph reading in millimeters of the earthquake recorded 100 kilometers from the epicenter. (a) Show that M (0.001) = 0. (b) Compute M (80, 000). (c) Show that an earthquake which registered 6.7 on the Richter scale had a seismograph reading ten times larger than one which measured 5.7. (d) Find two news stories about recent earthquakes which give their magnitudes on the Richter scale. How many times larger was the seismograph reading of the earthquake with larger magnitude? 76. While the decibel scale can be used in many disciplines,13 we shall restrict our attention to its use in acoustics, speciﬁcally its use in measuring the intensity level of sound.14 The Sound Intensity Level L (measured in decibels) of a sound intensity I (measured in watts per square meter) is given by L(I) = 10 log I
10−12. Like the Richter scale, this scale compares I to baseline: 10−12 W hearing. m2 is the threshold of human (a) Compute L(10−6). 12Rock-solid, perhaps? 13See this webpage for more information. 14As of the writing of this exercise, the Wikipedia page given here states that it may not meet the “general notability guideline” nor does it cite any references or sources. I ﬁnd this odd because it is this very usage of the decibel scale which shows up in every College Algebra book I have read. Perhaps those other books have been wrong all along and we’re just blindly following tradition. 432 Exponential and Logarithmic Functions (b) Damage to your hearing can start with short term exposure to sound levels around 115 decibels. What intensity I is needed to produce this level? (c) Compute L(1). How does this compare with the threshold of pain which is around 140 decibels? 77. The pH of a solution is a measure of its acidity or alkalinity. Speciﬁcally, pH = − log[H+] where [H+] is the hydrogen ion concentration in moles per liter. A solution with a pH less than 7 is an acid, one with a pH greater than 7 is a base (alkaline) and a pH of 7 is regarded as neutral. (a) The hydrogen ion concentration of pure water is [H+] = 10−7. Find its pH. (b) Find the pH of a solution with [H+] = 6.3 × 10−13. (c) The pH of gastric acid (the acid in your stomach) is about 0.7. What is the corresponding hydrogen ion concentration? 78. Show that logb 1 = 0 and logb b = 1 for every b > 0, b = 1. 79. (Crazy bonus question) Without using your calculator, determine which is larger: eπ or πe. 6.1 Introduction to Exponential and Logarithmic Functions 433 6.1.2 Answers 1. log2(8) = 3 4. log 1 3 (9) = −2 7. ln(1) = 0 10. 3−4 = 1 81 13. 10−1 = 0.1 2. log5 1 125 = −3 5. log 4 25 5 2 = − 1 2
8. 52 = 25 11. 4 3 −1 = 3 4 14. e1 = e 3. log4(32) = 5 2 6. log(0.001) = −3 9. (25) 1 2 = 5 12. 102 = 100 15. e− 1 2 = 1√ e 16. log3(27) = 3 17. log6(216) = 3 18. log2(32) = 5 19. log6 1 36 = −2 20. log8(4) = 2 3 21. log36(216) = 3 2 22. log 1 5 (625) = −4 23. log 1 6 (216) = −3 25. log 1 1000000 = −6 28. log4(8) = 3 2 31. log36 √ 4 36 = 1 4 26. log(0.01) = −2 29. log6(1) = 0 32. 7log7(3) = 3 34. log36 36216 = 216 35. ln(e5) = 5 37. log 3√ 105 = 5 3 40. log eln(100) = 2 38. ln 1√ e = − 1 2 24. log36(36) = 1 27. ln e3 = 3 30. log13 √ 13 = 1 2 33. 36log36(216) = 216 36. log 9√ 1011 = 11 9 39. log5 3log3 5 = 1 41. log2 3− log3(2) = −1 42. ln 426 log(1) = 0 43. (−∞, ∞) 44. (−2, ∞) 45. (5, ∞) 46. (−∞, −6) ∪ (−3, ∞) 47. (−2, −1) ∪ (1, ∞) 48. (−6, −3) ∪ (5, ∞) 49. (4, 7) 52. [1, ∞) 50. (5, ∞) 51. (−∞, ∞) 53. (−∞, −7) ∪ (1, ∞) 54. (13, ∞) 55. (0, 125) ∪ (125, ∞) 56. No domain 57. (−∞, −3) ∪ 1 2, 2 434 Exponential and Logarithmic
Functions 58. Domain of g: (−∞, ∞) Range of g: (−1, ∞) 59. Domain of g: (−∞, ∞) Range of g: (0, ∞) 3−2−1 x 1 2 3 H.A. y = −1 y = g(x) = 2x − 1 60. Domain of g: (−∞, ∞) Range of g: (2, ∞) y 11 10.A. y = 2 −3−2−1 1 2 3 x y = g(x) = 3−3−2−1 1 2 3 x y = g(x) = 1 3 x−1 61. Domain of g: (−∞, ∞) Range of g: (−20, ∞) y 80 70 60 50 40 30 20 10 −3−2 −10 1 2 3 x H.A. y = −20 y = g(x) = 10 x+1 2 − 20 62. Domain of g: (−∞, ∞) Range of g: (−∞, 8) 63. Domain of g: (−∞, ∞) Range of g: (0, ∞) H.A3−2−1 1 2 3 x y = g(x) = 8 − e−x y 80 70 60 50 40 30 20 10 −10 10 20 30 x y = g(x) = 10e−0.1x 6.1 Introduction to Exponential and Logarithmic Functions 435 64. Domain of g: (−1, ∞) Range of g: (−∞, ∞) 65. Domain of g: (0, ∞) Range of g: (−∞, ∞) 1 x −2 −3 V.A. x = −1 y = g(x) = log2(x + 1) y 3 2 1 −1 −2 −(x) = log 1 3 (x) + 1 66. Domain of g: (2, ∞) Range of g: (−∞, ∞) 67. Domain of g: (−20, ∞) Range of g: (−∞, ∞) y 3 2 1 −1 −2 − 10 11 x V.A. x = 2 y = g(x) = − log3(x − 2) y 3 2 1 −10 10 20 30 40 50 60
70 80 90 100 x −2 −3 V.A. x = −20 y = g(x) = 2 log(x + 20) − 1 68. Domain of g: (−∞, 8) Range of g:(−∞, ∞) y 3 2 1 −1 −2 −.A. x = 8 69. Domain of g: (0, ∞) Range of g: (−∞, ∞) y 30 20 10 −10 10 20 30 40 50 60 70 80 x y = g(x) = − ln(8 − x) y = g(x) = −10 ln x 10 436 Exponential and Logarithmic Functions 71. f (x) = 3x+2 − 4 f −1(x) = log3(x + 4) − 2 72. f (x) = log4(x − 1) f −1(x) = 4x + 1 y 6 5 4 3 2 1 −4−3−2−1 −1 1 2 3 4 5 6 x −2 −3 −4 y = f (x) = 3x+2 − 4 y = f −1(x) = log3(x + 4) − 2 73. f (x) = −2−x + 1 f −1(x) = − log2(1 − x) y 2 1 −2 −1 1 2 x −1 −2 y = f (x) = −2−x + 1 y = f −1(x) = − log2(1 − x) y 6 5 4 3 2 1 −2 −1 1 2 3 4 5 6 −1 −2 x y = f (x) = log4(x − 1) y = f −1(x) = 4x + 1 74. f (x) = 5 log(x) − 2 f −1(x) = 10 x+2 5 y 5 4 3 2 1 −4−3−2−1 −1 1 2 3 4 5 x −2 −3 −4 y = f (x) = 5 log(x) − 2 x+2 y = f −1(x) = 10 5 75. 76. 77. (a) M (0.001) = log 0.001 0.001 80,000 0.001 (b) M (80, 000) = log = log(1) = 0
. = log(80, 000, 000) ≈ 7.9. (a) L(10−6) = 60 decibels. (b) I = 10−.5 ≈ 0.316 watts per square meter. (c) Since L(1) = 120 decibels and L(100) = 140 decibels, a sound with intensity level 140 decibels has an intensity 100 times greater than a sound with intensity level 120 decibels. (a) The pH of pure water is 7. (b) If [H+] = 6.3 × 10−13 then the solution has a pH of 12.2. (c) [H+] = 10−0.7 ≈.1995 moles per liter. 6.2 Properties of Logarithms 437 6.2 Properties of Logarithms In Section 6.1, we introduced the logarithmic functions as inverses of exponential functions and discussed a few of their functional properties from that perspective. In this section, we explore the algebraic properties of logarithms. Historically, these have played a huge role in the scientiﬁc development of our society since, among other things, they were used to develop analog computing devices called slide rules which enabled scientists and engineers to perform accurate calculations leading to such things as space travel and the moon landing. As we shall see shortly, logs inherit analogs of all of the properties of exponents you learned in Elementary and Intermediate Algebra. We ﬁrst extract two properties from Theorem 6.2 to remind us of the deﬁnition of a logarithm as the inverse of an exponential function. Theorem 6.3. (Inverse Properties of Exponential and Logarithmic Functions) Let b > 0, b = 1. ba = c if and only if logb(c) = a logb (bx) = x for all x and blogb(x) = x for all x > 0 Next, we spell out what it means for exponential and logarithmic functions to be one-to-one. Theorem 6.4. (One-to-one Properties of Exponential and Logarithmic Functions) Let f (x) = bx and g(x) = logb(x) where b > 0, b = 1. Then f and g are one-to-one and bu =
bw if and only if u = w for all real numbers u and w. logb(u) = logb(w) if and only if u = w for all real numbers u > 0, w > 0. We now state the algebraic properties of exponential functions which will serve as a basis for the properties of logarithms. While these properties may look identical to the ones you learned in Elementary and Intermediate Algebra, they apply to real number exponents, not just rational exponents. Note that in the theorem that follows, we are interested in the properties of exponential functions, so the base b is restricted to b > 0, b = 1. An added beneﬁt of this restriction is that it eliminates the pathologies discussed in Section 5.3 when, for example, we simpliﬁed x2/33/2 and obtained \|x\| instead of what we had expected from the arithmetic in the exponents, x1 = x. Theorem 6.5. (Algebraic Properties of Exponential Functions) Let f (x) = bx be an exponential function (b > 0, b = 1) and let u and w be real numbers. Product Rule: f (u + w) = f (u)f (w). In other words, bu+w = bubw Quotient Rule: f (u − w) = f (u) f (w). In other words, bu−w = bu bw Power Rule: (f (u))w = f (uw). In other words, (bu)w = buw While the properties listed in Theorem 6.5 are certainly believable based on similar properties of integer and rational exponents, the full proofs require Calculus. To each of these properties of 438 Exponential and Logarithmic Functions exponential functions corresponds an analogous property of logarithmic functions. We list these below in our next theorem. Theorem 6.6. (Algebraic Properties of Logarithmic Functions) Let g(x) = logb(x) be a logarithmic function (b > 0, b = 1) and let u > 0 and w > 0 be real numbers. Product Rule: g(uw) = g(u) + g(w). In other words, logb(uw) = logb(u) + logb(w) Quotient Rule: g u w = g
(u) − g(w). In other words, logb u w = logb(u) − logb(w) Power Rule: g (uw) = wg(u). In other words, logb (uw) = w logb(u) There are a couple of diﬀerent ways to understand why Theorem 6.6 is true. Consider the product rule: logb(uw) = logb(u) + logb(w). Let a = logb(uw), c = logb(u), and d = logb(w). Then, by deﬁnition, ba = uw, bc = u and bd = w. Hence, ba = uw = bcbd = bc+d, so that ba = bc+d. By the one-to-one property of bx, we have a = c + d. In other words, logb(uw) = logb(u) + logb(w). The remaining properties are proved similarly. From a purely functional approach, we can see the properties in Theorem 6.6 as an example of how inverse functions interchange the roles of inputs in outputs. For instance, the Product Rule for exponential functions given in Theorem 6.5, f (u + w) = f (u)f (w), says that adding inputs results in multiplying outputs. Hence, whatever f −1 is, it must take the products of outputs from f and return them to the sum of their respective inputs. Since the outputs from f are the inputs to f −1 and vice-versa, we have that that f −1 must take products of its inputs to the sum of their respective outputs. This is precisely what the Product Rule for Logarithmic functions states in Theorem 6.6: g(uw) = g(u) + g(w). The reader is encouraged to view the remaining properties listed in Theorem 6.6 similarly. The following examples help build familiarity with these properties. In our ﬁrst example, we are asked to ‘expand’ the logarithms. This means that we read the properties in Theorem 6.6 from left to right and rewrite products inside the log as sums outside the log, quotients inside the log as diﬀerences outside the log, and powers inside the log as factors outside the log.1 Example 6.
2.1. Expand the following using the properties of logarithms and simplify. Assume when necessary that all quantities represent positive real numbers. 1. log2 8 x 4. log 3 100x2 yz5 Solution. 2. log0.1 10x2 5. log117 x2 − 4 3. ln 2 3 ex 1. To expand log2 8 x, we use the Quotient Rule identifying u = 8 and w = x and simplify. 1Interestingly enough, it is the exact opposite process (which we will practice later) that is most useful in Algebra, the utility of expanding logarithms becomes apparent in Calculus. 6.2 Properties of Logarithms 439 log2 8 x = log2(8) − log2(x) Quotient Rule = 3 − log2(x) = − log2(x) + 3 Since 23 = 8 2. In the expression log0.1 10x2, we have a power (the x2) and a product. In order to use the Product Rule, the entire quantity inside the logarithm must be raised to the same exponent. Since the exponent 2 applies only to the x, we ﬁrst apply the Product Rule with u = 10 and w = x2. Once we get the x2 by itself inside the log, we may apply the Power Rule with u = x and w = 2 and simplify. log0.1 10x2 = log0.1(10) + log0.1 x2 Product Rule = log0.1(10) + 2 log0.1(x) = −1 + 2 log0.1(x) = 2 log0.1(x) − 1 Power Rule Since (0.1)−1 = 10 3. We have a power, quotient and product occurring in ln 3 ex. Since the exponent 2 applies to the entire quantity inside the logarithm, we begin with the Power Rule with u = 3 ex and w = 2. Next, we see the Quotient Rule is applicable, with u = 3 and w = ex, so we replace is being multiplied by 2, the entire ln 3 ex quantity ln(3) − ln(ex) is multiplied by 2. Finally, we apply the Product Rule with u = e and w = x, and replace ln(ex) with the quantity ln(e) +
ln(x), and simplify, keeping in mind that the natural log is log base e. with the quantity ln(3) − ln(ex). Since ln 3 ex 2 2 3 ex ln = 2 ln 3 ex Power Rule = 2 [ln(3) − ln(ex)] Quotient Rule = 2 ln(3) − 2 ln(ex) = 2 ln(3) − 2 [ln(e) + ln(x)] Product Rule = 2 ln(3) − 2 ln(e) − 2 ln(x) = 2 ln(3) − 2 − 2 ln(x) Since e1 = e = −2 ln(x) + 2 ln(3) − 2 4. In Theorem 6.6, there is no mention of how to deal with radicals. However, thinking back to 3 exponent. We begin by using the Power Deﬁnition 5.5, we can rewrite the cube root as a 1 440 Exponential and Logarithmic Functions Rule2, and we keep in mind that the common log is log base 10. log 3 100x2 yz5 = log 1/3 100x2 yz5 100x2 yz5 = 1 3 log log 100x2 − log yz5 3 log yz5 3 log 100x2 − 1 log(100) + log x2 − 1 3 3 log x2 − 1 3 log(100) + 1 3 log(x) − 1 3 log(100) + 2 3 + 2 3 log(x) − 1 3 log(x) − 1 3 log(y) − 5 3 log(y log(y) + log z5 3 log z5 3 log(y) − 1 3 log(z) 3 log(y) − 5 3 log(z) 3 log(z) + 2 3 Power Rule Quotient Rule Product Rule Power Rule Since 102 = 100 5. At ﬁrst it seems as if we have no means of simplifying log117 x2 − 4, since none of the properties of logs addresses the issue of expanding a diﬀerence inside the logarithm. However, we may factor x2 − 4 = (x + 2)(x − 2) thereby introducing a product which gives us license to use the Product Rule. log117 x2 − 4 = log117
[(x + 2)(x − 2)] Factor = log117(x + 2) + log117(x − 2) Product Rule A couple of remarks about Example 6.2.1 are in order. First, while not explicitly stated in the above example, a general rule of thumb to determine which log property to apply ﬁrst to a complicated problem is ‘reverse order of operations.’ For example, if we were to substitute a number for x into 10x2, we would ﬁrst square the x, then multiply by 10. The last step is the the expression log0.1 multiplication, which tells us the ﬁrst log property to apply is the Product Rule. In a multi-step problem, this rule can give the required guidance on which log property to apply at each step. The reader is encouraged to look through the solutions to Example 6.2.1 to see this rule in action. Second, while we were instructed to assume when necessary that all quantities represented positive real numbers, the authors would be committing a sin of omission if we failed to point out that, for x2 − 4 and g(x) = log117(x + 2) + log117(x − 2) have diﬀerent instance, the functions f (x) = log117 domains, and, hence, are diﬀerent functions. We leave it to the reader to verify the domain of f is (−∞, −2) ∪ (2, ∞) whereas the domain of g is (2, ∞). In general, when using log properties to 2At this point in the text, the reader is encouraged to carefully read through each step and think of which quantity is playing the role of u and which is playing the role of w as we apply each property. 6.2 Properties of Logarithms 441 expand a logarithm, we may very well be restricting the domain as we do so. One last comment before we move to reassembling logs from their various bits and pieces. The authors are well aware of the propensity for some students to become overexcited and invent their own properties of logs x2 − log117(4), which simply isn’t true, in general. The unwritten3 like log117 property of logarithms is that if it isn’t written in a textbook, it probably isn’t true. x2 − 4 = log117
Example 6.2.2. Use the properties of logarithms to write the following as a single logarithm. 1. log3(x − 1) − log3(x + 1) 2. log(x) + 2 log(y) − log(z) 3. 4 log2(x) + 3 4. − ln(x) − 1 2 Solution. Whereas in Example 6.2.1 we read the properties in Theorem 6.6 from left to right to expand logarithms, in this example we read them from right to left. 1. The diﬀerence of logarithms requires the Quotient Rule: log3(x−1)−log3(x+1) = log3 x−1 x+1. 2. In the expression, log(x) + 2 log(y) − log(z), we have both a sum and diﬀerence of logarithms. However, before we use the product rule to combine log(x) + 2 log(y), we note that we need to somehow deal with the coeﬃcient 2 on log(y). This can be handled using the Power Rule. We can then apply the Product and Quotient Rules as we move from left to right. Putting it all together, we have log(x) + 2 log(y) − log(z) = log(x) + log y2 − log(z) Power Rule = log xy2 − log(z) Product Rule = log xy2 z Quotient Rule 3. We can certainly get started rewriting 4 log2(x) + 3 by applying the Power Rule to 4 log2(x) x4, but in order to use the Product Rule to handle the addition, we need to to obtain log2 rewrite 3 as a logarithm base 2. From Theorem 6.3, we know 3 = log2 23, so we get 4 log2(x) + 3 = log2 = log2 = log2 = log2 x4 + 3 x4 + log2 x4 + log2(8) 8x4 23 Since 3 = log2 Power Rule 23 Product Rule 3The authors relish the irony involved in writing what follows. 442 Exponential and Logarithmic Functions 4. To get started with − ln(x) − 1 2, we rewrite −
ln(x) as (−1) ln(x). We can then use the Power Rule to obtain (−1) ln(x) = ln x−1. In order to use the Quotient Rule, we need to write 1 2 as a natural logarithm. Theorem 6.3 gives us 1 e). We have 2 = ln e1/2 = ln ( √ − ln(x) − 1 2 2 = (−1) ln(x) − 1 = ln x−1 − 1 2 = ln x−1 − ln e1/2 Since 1 = ln x−1 − ln ( x−1 √ e 1 √ x = ln = ln e) √ e Power Rule 2 = ln e1/2 Quotient Rule As we would expect, the rule of thumb for re-assembling logarithms is the opposite of what it was for dismantling them. That is, if we are interested in rewriting an expression as a single logarithm, we apply log properties following the usual order of operations: deal with multiples of logs ﬁrst with the Power Rule, then deal with addition and subtraction using the Product and Quotient Rules, respectively. Additionally, we ﬁnd that using log properties in this fashion can increase the domain of the expression. For example, we leave it to the reader to verify the domain of f (x) = log3(x−1)−log3(x+1) is (1, ∞) but the domain of g(x) = log3 is (−∞, −1)∪(1, ∞). We will need to keep this in mind when we solve equations involving logarithms in Section 6.4 - it is precisely for this reason we will have to check for extraneous solutions. x−1 x+1 The two logarithm buttons commonly found on calculators are the ‘LOG’ and ‘LN’ buttons which correspond to the common and natural logs, respectively. Suppose we wanted an approximation to log2(7). The answer should be a little less than 3, (Can you explain why?) but how do we coerce the calculator into telling us a more accurate answer? We need the following theorem. Theorem 6.7. (Change of Base Formulas) Let a, b >
0, a, b = 1. ax = bx logb(a) for all real numbers x. loga(x) = logb(x) logb(a) for all real numbers x > 0. The proofs of the Change of Base formulas are a result of the other properties studied in this section. If we start with bx logb(a) and use the Power Rule in the exponent to rewrite x logb(a) as logb (ax) and then apply one of the Inverse Properties in Theorem 6.3, we get bx logb(a) = blogb(ax) = ax, 6.2 Properties of Logarithms 443 as required. To verify the logarithmic form of the property, we also use the Power Rule and an Inverse Property. We note that loga(x) · logb(a) = logb aloga(x) = logb(x), and we get the result by dividing through by logb(a). Of course, the authors can’t help but point out the inverse relationship between these two change of base formulas. To change the base of an exponential expression, we multiply the input by the factor logb(a). To change the base of a logarithmic expression, we divide the output by the factor logb(a). While, in the grand scheme of things, both change of base formulas are really saying the same thing, the logarithmic form is the one usually encountered in Algebra while the exponential form isn’t usually introduced until Calculus.4 What Theorem 6.7 really tells us is that all exponential and logarithmic functions are just scalings of one another. Not only does this explain why their graphs have similar shapes, but it also tells us that we could do all of mathematics with a single base - be it 10, e, 42, or 117. Your Calculus teacher will have more to say about this when the time comes. Example 6.2.3. Use an appropriate change of base formula to convert the following expressions to ones with the indicated base. Verify your answers using a calculator, as appropriate. 1. 32 to base 10 3. log4(5) to base e Solution. 2. 2x to base e 4. ln(x) to base 10 1. We apply the Change of Base formula with a = 3 and b = 10 to obtain 32 = 102 log(3
). Typing the latter in the calculator produces an answer of 9 as required. 2. Here, a = 2 and b = e so we have 2x = ex ln(2). To verify this on our calculator, we can graph f (x) = 2x and g(x) = ex ln(2). Their graphs are indistinguishable which provides evidence that they are the same function. 4The authors feel so strongly about showing students that every property of logarithms comes from and corresponds to a property of exponents that we have broken tradition with the vast majority of other authors in this ﬁeld. This isn’t the ﬁrst time this happened, and it certainly won’t be the last. y = f (x) = 2x and y = g(x) = ex ln(2) 444 Exponential and Logarithmic Functions 3. Applying the change of base with a = 4 and b = e leads us to write log4(5) = ln(5) this in the calculator gives ln(5) By deﬁnition, log4(5) is the exponent we put on 4 to get 5. The calculator conﬁrms this.5 ln(4). Evaluating ln(4) ≈ 1.16. How do we check this really is the value of log4(5)? 4. We write ln(x) = loge(x) = log(x) both graphs appear to be identical. log(e). We graph both f (x) = ln(x) and g(x) = log(x) log(e) and ﬁnd y = f (x) = ln(x) and y = g(x) = log(x) log(e) 5Which means if it is lying to us about the ﬁrst answer it gave us, at least it is being consistent. 6.2 Properties of Logarithms 445 6.2.1 Exercises In Exercises 1 - 15, expand the given logarithm and simplify. Assume when necessary that all quantities represent positive real numbers. 1. ln(x3y2) 4. log(1.23 × 1037) 7. log√ 2 4x3 10. log3 x2 81y4 13. log √ 100x
√ 3 10 y 2. log2 128 x2 + 4 √ z xy 5. ln 8. log 1 3 (9x(y3 − 8)) 11. ln 4 xy ez 4 3√ √ y x2 z 14. log 1 2 3. log5 3 z 25 6. log5 x2 − 25 9. log 1000x3y5 12. log6 4 216 x3y 15. ln √ 3 x √ yz 10 In Exercises 16 - 29, use the properties of logarithms to write the expression as a single logarithm. 16. 4 ln(x) + 2 ln(y) 17. log2(x) + log2(y) − log2(z) 18. log3(x) − 2 log3(y) 19. 1 2 log3(x) − 2 log3(y) − log3(z) 20. 2 ln(x) − 3 ln(y) − 4 ln(z) 22. − 1 3 ln(x) − 1 3 ln(y) + 1 3 ln(z) 24. 3 − log(x) 26. ln(x) + 1 2 21. log(x) − 1 3 log(z) + 1 2 log(y) 23. log5(x) − 3 25. log7(x) + log7(x − 3) − 2 27. log2(x) + log4(x) 28. log2(x) + log4(x − 1) 29. log2(x) + log 1 2 (x − 1) 446 Exponential and Logarithmic Functions In Exercises 30 - 33, use the appropriate change of base formula to convert the given expression to an expression with the indicated base. 30. 7x−1 to base e x 32. to base e 2 3 31. log3(x + 2) to base 10 33. log(x2 + 1) to base e In Exercises 34 - 39, use the appropriate change of base formula to approximate the logarithm. 34. log3(12) 1 10 37. log4 35. log5(80) 38. log 3 5 (1000) 36. log6(72) 39. log 2 3 (50) 40. Compare and contrast the
graphs of y = ln(x2) and y = 2 ln(x). 41. Prove the Quotient Rule and Power Rule for Logarithms. 42. Give numerical examples to show that, in general, (a) logb(x + y) = logb(x) + logb(y) (b) logb(x − y) = logb(x) − logb(y) (c) logb x y = logb(x) logb(y) 43. The Henderson-Hasselbalch Equation: Suppose HA represents a weak acid. Then we have a reversible chemical reaction HA H + + A−. The acid disassociation constant, Ka, is given by Kα = [H +][A−] [HA] = [H +] [A−] [HA], where the square brackets denote the concentrations just as they did in Exercise 77 in Section 6.1. The symbol pKa is deﬁned similarly to pH in that pKa = − log(Ka). Using the deﬁnition of pH from Exercise 77 and the properties of logarithms, derive the Henderson-Hasselbalch Equation which states pH = pKa + log [A−] [HA] 44. Research the history of logarithms including the origin of the word ‘logarithm’ itself. Why is the abbreviation of natural log ‘ln’ and not ‘nl’? 45. There is a scene in the movie ‘Apollo 13’ in which several people at Mission Control use slide rules to verify a computation. Was that scene accurate? Look for other pop culture references to logarithms and slide rules. 6.2 Properties of Logarithms 447 6.2.2 Answers 1. 3 ln(x) + 2 ln(y) 3. 3 log5(z) − 6 5. 1 2 ln(z) − ln(x) − ln(y) 7. 3 log√ 2(x) + 4 2. 7 − log2(x2 + 4) 4. log(1.23) + 37 6. log5(x − 5) + log5(x + 5) 8. −2 + log 1 3 (x) + log 1 3 (y − 2) + log 1 3 (y2
+ 2y + 4) 9. 3 + 3 log(x) + 5 log(y) 10. 2 log3(x) − 4 − 4 log3(y) 11. 1 4 ln(x) + 1 4 ln(y) − 1 4 − 1 4 ln(z) 12. 12 − 12 log6(x) − 4 log6(y) 13. 5 3 + log(x) + 1 2 log(y) 14. −2 + 2 3 log 1 2 (x) − log 1 2 (y) − 1 2 log 1 2 (z) 15. 1 3 ln(x) − ln(10) − 1 17. log2 xy z 2 ln(y) − 1 2 ln(z) 20. ln x2 y3z4 23. log5 x 125 26. ln (x √ e) 29. log2 x x−1 32. 2 3 x = ex ln( 2 3 ) 34. log3(12) ≈ 2.26186 36. log6(72) ≈ 2.38685 38. log 3 5 (1000) ≈ −13.52273 18. log3 x y2 21. log √ x 3√ y z 24. log 1000 x 27. log2 x3/2 16. ln(x4y2) 19. log3 √ x y2z 22. ln z xy 3 25. log7 28. log2 x(x−3) 49 √ x x − 1 30. 7x−1 = e(x−1) ln(7) 31. log3(x + 2) = log(x+2) log(3) 33. log(x2 + 1) = ln(x2+1) ln(10) 35. log5(80) ≈ 2.72271 37. log4 1 10 ≈ −1.66096 39. log 2 3 (50) ≈ −9.64824 448 Exponential and Logarithmic Functions 6.3 Exponential Equations and Inequalities In this section we will develop techniques for solving equations involving exponential functions. Suppose, for instance, we wanted to solve the equation 2x = 128. After a moment’s calculation, we ﬁnd 128 = 27, so we have 2x = 27
. The one-to-one property of exponential functions, detailed in Theorem 6.4, tells us that 2x = 27 if and only if x = 7. This means that not only is x = 7 a solution to 2x = 27, it is the only solution. Now suppose we change the problem ever so slightly to 2x = 129. We could use one of the inverse properties of exponentials and logarithms listed in Theorem 6.3 to write 129 = 2log2(129). We’d then have 2x = 2log2(129), which means our solution is x = log2(129). This makes sense because, after all, the deﬁnition of log2(129) is ‘the exponent we put on 2 to get 129.’ Indeed we could have obtained this solution directly by rewriting the equation 2x = 129 in its logarithmic form log2(129) = x. Either way, in order to get a reasonable decimal approximation to this number, we’d use the change of base formula, Theorem 6.7, to give us something more calculator friendly,1 say log2(129) = ln(129) ln(2). Another way to arrive at this answer is as follows 2x = 129 ln (2x) = ln(129) Take the natural log of both sides. x ln(2) = ln(129) Power Rule ln(129) ln(2) x = ‘Taking the natural log’ of both sides is akin to squaring both sides: since f (x) = ln(x) is a function, as long as two quantities are equal, their natural logs are equal.2 Also note that we treat ln(2) as any other non-zero real number and divide it through3 to isolate the variable x. We summarize below the two common ways to solve exponential equations, motivated by our examples. Steps for Solving an Equation involving Exponential Functions 1. Isolate the exponential function. 2. (a) If convenient, express both sides with a common base and equate the exponents. (b) Otherwise, take the natural log of both sides of the equation and use the Power Rule. Example 6.3.1. Solve the following equations. Check your answer graphically using a calculator. 1. 23x = 161−x 4. 75 = 100
1+3e−2t Solution. 2. 2000 = 1000 · 3−0.1t 3. 9 · 3x = 72x 5. 25x = 5x + 6 6. ex−e−x 2 = 5 1You can use natural logs or common logs. We choose natural logs. (In Calculus, you’ll learn these are the most ‘mathy’ of the logarithms.) 2This is also the ‘if’ part of the statement logb(u) = logb(w) if and only if u = w in Theorem 6.4. 3Please resist the temptation to divide both sides by ‘ln’ instead of ln(2). Just like it wouldn’t make sense to √ √ divide both sides by the square root symbol ‘ ’ when solving x 2 = 5, it makes no sense to divide by ‘ln’. 6.3 Exponential Equations and Inequalities 449 1. Since 16 is a power of 2, we can rewrite 23x = 161−x as 23x = 241−x. Using properties of exponents, we get 23x = 24(1−x). Using the one-to-one property of exponential functions, we get 3x = 4(1−x) which gives x = 4 7. To check graphically, we set f (x) = 23x and g(x) = 161−x and see that they intersect at x = 4 7 ≈ 0.5714. 2. We begin solving 2000 = 1000 · 3−0.1t by dividing both sides by 1000 to isolate the exponential which yields 3−0.1t = 2. Since it is inconvenient to write 2 as a power of 3, we use the natural log to get ln 3−0.1t = ln(2). Using the Power Rule, we get −0.1t ln(3) = ln(2), so we divide both sides by −0.1 ln(3) to get t = − ln(2) ln(3). On the calculator, we graph f (x) = 2000 and g(x) = 1000 · 3−0.1x and ﬁnd that they intersect at x = − 10 ln(2) ln(3) ≈ −6.3093. 0.1 ln
(3) = − 10 ln(2) y = f (x) = 23x and y = g(x) = 161−x y = f (x) = 2000 and y = g(x) = 1000 · 3−0.1x 3. We ﬁrst note that we can rewrite the equation 9·3x = 72x as 32 ·3x = 72x to obtain 3x+2 = 72x. Since it is not convenient to express both sides as a power of 3 (or 7 for that matter) we use ln 3x+2 = ln 72x. The power rule gives (x + 2) ln(3) = 2x ln(7). Even the natural log: though this equation appears very complicated, keep in mind that ln(3) and ln(7) are just constants. The equation (x + 2) ln(3) = 2x ln(7) is actually a linear equation and as such we gather all of the terms with x on one side, and the constants on the other. We then divide both sides by the coeﬃcient of x, which we obtain by factoring. (x + 2) ln(3) = 2x ln(7) x ln(3) + 2 ln(3) = 2x ln(7) 2 ln(3) = 2x ln(7) − x ln(3) 2 ln(3) = x(2 ln(7) − ln(3)) Factor. x = 2 ln(3) 2 ln(7)−ln(3) Graphing f (x) = 9·3x and g(x) = 72x on the calculator, we see that these two graphs intersect at x = 2 ln(3) 2 ln(7)−ln(3) ≈ 0.7866. 4. Our objective in solving 75 = 100 is to ﬁrst isolate the exponential. To that end, we clear denominators and get 75 1 + 3e−2t = 100. From this we get 75 + 225e−2t = 100, which leads to 225e−2t = 25, and ﬁnally, e−2t = 1 9. Taking the natural log of both sides 1+3e−2t 450 Exponential
and Logarithmic Functions. Since natural log is log base e, ln e−2t = −2t. We can also use gives ln e−2t = ln 1 9 = − ln(9). Putting these two steps together, we simplify the Power Rule to write ln 1 9 to −2t = − ln(9). We arrive at our solution, t = ln(9) ln e−2t = ln 1 2 which simpliﬁes to 9 t = ln(3). (Can you explain why?) The calculator conﬁrms the graphs of f (x) = 75 and g(x) = 100 1+3e−2x intersect at x = ln(3) ≈ 1.099. y = f (x) = 9 · 3x and y = g(x) = 72x y = f (x) = 75 and y = g(x) = 100 1+3e−2x 5. We start solving 25x = 5x + 6 by rewriting 25 = 52 so that we have 52x = 5x + 6, or 52x = 5x + 6. Even though we have a common base, having two terms on the right hand side of the equation foils our plan of equating exponents or taking logs. If we stare at this long enough, we notice that we have three terms with the exponent on one term exactly twice that of another. To our surprise and delight, we have a ‘quadratic in disguise’. Letting u = 5x, we have u2 = (5x)2 = 52x so the equation 52x = 5x + 6 becomes u2 = u + 6. Solving this as u2 − u − 6 = 0 gives u = −2 or u = 3. Since u = 5x, we have 5x = −2 or 5x = 3. Since 5x = −2 has no real solution, (Why not?) we focus on 5x = 3. Since it isn’t convenient to express 3 as a power of 5, we take natural logs and get ln (5x) = ln(3) so that x ln(5) = ln(3) or x = ln(3) ln(5). On the calculator, we see the graphs of f (x) = 25x
and g(x) = 5x + 6 intersect at x = ln(3) ln(5) ≈ 0.6826. 2 6. At ﬁrst, it’s unclear how to proceed with ex−e−x = 5, besides clearing the denominator to obtain ex − e−x = 10. Of course, if we rewrite e−x = 1 ex, we see we have another denominator lurking in the problem: ex − 1 ex = 10. Clearing this denominator gives us e2x − 1 = 10ex, and once again, we have an equation with three terms where the exponent on one term is exactly twice that of another - a ‘quadratic in disguise.’ If we let u = ex, then u2 = e2x so the equation e2x − 1 = 10ex can be viewed as u2 − 1 = 10u. Solving u2 − 10u − 1 = 0, we obtain 26. Since 5 − by the quadratic formula u = 5 ± 26 < 0, we get no real solution to ex = 5 − 26, we take natural logs to obtain x = ln 5 + and g(x) = 5, we see that the graphs intersect at x = ln 5 + 26. From this, we have ex = 5 ± √ 26, but for ex = 5 + 26. If we graph f (x) = ex−e−x √ 26 ≈ 2.312 √ √ √ √ √ 2 6.3 Exponential Equations and Inequalities 451 y = f (x) = 25x and y = g(x) = 5x + 6 y = f (x) = ex−e−x y = g(x) = 5 2 and The authors would be remiss not to mention that Example 6.3.1 still holds great educational value. Much can be learned about logarithms and exponentials by verifying the solutions obtained in Example 6.3.1 analytically. For example, to verify our solution to 2000 = 1000 · 3−0.1t, we substitute t = − 10 ln(2) ln(3) and obtain 2000 2000 2000 2000 2000 − 10 ln(2) ln(3) −0.1 ln(2) ln(3)? = 1000 · 3? = 1000 · 3?
= 1000 · 3log3(2)? = 1000 · 2 = 2000 Change of Base Inverse Property The other solutions can be veriﬁed by using a combination of log and inverse properties. Some fall out quite quickly, while others are more involved. We leave them to the reader. Since exponential functions are continuous on their domains, the Intermediate Value Theorem 3.1 applies. As with the algebraic functions in Section 5.3, this allows us to solve inequalities using sign diagrams as demonstrated below. Example 6.3.2. Solve the following inequalities. Check your answer graphically using a calculator. 1. 2x2−3x − 16 ≥ 0 Solution. 2. ex ex − 4 ≤ 3 3. xe2x < 4x 1. Since we already have 0 on one side of the inequality, we set r(x) = 2x2−3x − 16. The domain of r is all real numbers, so in order to construct our sign diagram, we need to ﬁnd the zeros of r. Setting r(x) = 0 gives 2x2−3x − 16 = 0 or 2x2−3x = 16. Since 16 = 24 we have 2x2−3x = 24, so by the one-to-one property of exponential functions, x2 − 3x = 4. Solving x2 − 3x − 4 = 0 gives x = 4 and x = −1. From the sign diagram, we see r(x) ≥ 0 on (−∞, −1] ∪ [4, ∞), which corresponds to where the graph of y = r(x) = 2x2−3x − 16, is on or above the x-axis. 452 Exponential and Logarithmic Functions (+) 0 (−) 0 (+) −1 4 y = r(x) = 2x2−3x − 16 2. The ﬁrst step we need to take to solve ex ex−4 ≤ 3 is to get 0 on one side of the inequality. To that end, we subtract 3 from both sides and get a common denominator ex ex − 4 ≤ 3 ex ex − 4 − − 3 ≤ 0 ex ex − 4 3 (ex − 4) ex − 4 12 − 2ex ex − 4 ≤ 0 Common denomintors. ≤ 0 We set r(x) = 12−2ex ex−4 and we
note that r is undeﬁned when its denominator ex − 4 = 0, or when ex = 4. Solving this gives x = ln(4), so the domain of r is (−∞, ln(4)) ∪ (ln(4), ∞). To ﬁnd the zeros of r, we solve r(x) = 0 and obtain 12 − 2ex = 0. Solving for ex, we ﬁnd ex = 6, or x = ln(6). When we build our sign diagram, ﬁnding test values may be a little tricky since we need to check values around ln(4) and ln(6). Recall that the function ln(x) is increasing4 which means ln(3) < ln(4) < ln(5) < ln(6) < ln(7). While the prospect of determining the sign of r (ln(3)) may be very unsettling, remember that eln(3) = 3, so r (ln(3)) = 12 − 2eln(3) eln(3) − 4 = 12 − 2(3) 3 − 4 = −6 We determine the signs of r (ln(5)) and r (ln(7)) similarly.5 From the sign diagram, we ﬁnd our answer to be (−∞, ln(4)) ∪ [ln(6), ∞). Using the calculator, we see the graph of f (x) = ex ex−4 is below the graph of g(x) = 3 on (−∞, ln(4)) ∪ (ln(6), ∞), and they intersect at x = ln(6) ≈ 1.792. 4This is because the base of ln(x) is e > 1. If the base b were in the interval 0 < b < 1, then logb(x) would decreasing. 5We could, of course, use the calculator, but what fun would that be? 6.3 Exponential Equations and Inequalities 453 (−) (+) 0 (−) ln(4) ln(6) y = f (x) = ex ex−4 y = g(x) = 3 3. As before, we start solving xe2x < 4x by getting 0 on one side of the inequality, xe2x
− 4x < 0. We set r(x) = xe2x − 4x and since there are no denominators, even-indexed radicals, or logs, the domain of r is all real numbers. Setting r(x) = 0 produces xe2x − 4x = 0. We factor to get x e2x − 4 = 0 which gives x = 0 or e2x − 4 = 0. To solve the latter, we isolate the exponential and take logs to get 2x = ln(4), or x = ln(4) 2 = ln(2). (Can you explain the last equality using properties of logs?) As in the previous example, we need to be careful about choosing test values. Since ln(1) = 0, we choose ln 1 and ln(3). Evaluating,6 we get 2, ln 3 2 r ln 1 2 = ln 1 2 = ln 1 2 = ln 1 2 4 ln 1 = 1 2 ) − 4 ln 1 e2 ln( 1 2 2 )2 − 4 ln 1 eln( 1 2 eln( 1 4 ) − 4 ln 1 2 = − 15 − 4 ln 1 2 2 4 ln 1 2 Power Rule 2 < 1, ln 1 ) is (+), so r(x) < 0 on (0, ln(2)). The calculator Since 1 conﬁrms that the graph of f (x) = xe2x is below the graph of g(x) = 4x on these intervals.7 < 0 and we get r(ln 1 2 2 (+) 0 (−) 0 (+) 0 ln(2) y = f (x) = xe2x and y = g(x) = 4x 6A calculator can be used at this point. As usual, we proceed without apologies, with the analytical method. 7Note: ln(2) ≈ 0.693. 454 Exponential and Logarithmic Functions Example 6.3.3. Recall from Example 6.1.2 that the temperature of coﬀee T (in degrees Fahrenheit) t minutes after it is served can be modeled by T (t) = 70 + 90e−0.1t. When will the coﬀee be warmer than 100◦F? 3 so that t = −10 ln 1
Solution. We need to ﬁnd when T (t) > 100, or in other words, we need to solve the inequality 70 + 90e−0.1t > 100. Getting 0 on one side of the inequality, we have 90e−0.1t − 30 > 0, and we set r(t) = 90e−0.1t − 30. The domain of r is artiﬁcially restricted due to the context of the problem to [0, ∞), so we proceed to ﬁnd the zeros of r. Solving 90e−0.1t − 30 = 0 results in which, after a quick application of the Power Rule leaves us with e−0.1t = 1 t = 10 ln(3). If we wish to avoid using the calculator to choose test values, we note that since 1 < 3, 0 = ln(1) < ln(3) so that 10 ln(3) > 0. So we choose t = 0 as a test value in [0, 10 ln(3)). Since 3 < 4, 10 ln(3) < 10 ln(4), so the latter is our choice of a test value for the interval (10 ln(3), ∞). Our sign diagram is below, and next to it is our graph of y = T (t) from Example 6.1.2 with the horizontal line y = 100. 3 (+) 0 0 (−) 10 ln(3) y 180 160 140 120 80 60 40 20 y = 100 H.A. y = 70 2 4 6 8 10 12 14 16 18 20 t y = T (t) In order to interpret what this means in the context of the real world, we need a reasonable approximation of the number 10 ln(3) ≈ 10.986. This means it takes approximately 11 minutes for the coﬀee to cool to 100◦F. Until then, the coﬀee is warmer than that.8 We close this section by ﬁnding the inverse of a function which is a composition of a rational function with an exponential function. Example 6.3.4. The function f (x) = 5ex ex + 1 your answer graphically using your calculator. Solution. We start by writing y = f (x), and interchange the roles of x and y. To solve for y, we �
�rst clear denominators and then isolate the exponential function. is one-to-one. Find a formula for f −1(x) and check 8Critics may point out that since we needed to use the calculator to interpret our answer anyway, why not use it earlier to simplify the computations? It is a fair question which we answer unfairly: it’s our book. 6.3 Exponential Equations and Inequalities 455 Switch x and y y = x = 5ex ex + 1 5ey ey + 1 x (ey + 1) = 5ey xey + x = 5ey x = 5ey − xey x = ey(5 − x) ey = x 5 − x ln (ey) = ln y = ln −x We claim f −1(x) = ln. To verify this analytically, we would need to verify the compositions f −1 ◦ f (x) = x for all x in the domain of f and that f ◦ f −1 (x) = x for all x in the domain of f −1. We leave this to the reader. To verify our solution graphically, we graph y = f (x) = 5ex ex+1 and y = g(x) = ln on the same set of axes and observe the symmetry about the line y = x. Note the domain of f is the range of g and vice-versa. x 5−x y = f (x) = 5ex ex+1 and y = g(x) = ln x 5−x 456 Exponential and Logarithmic Functions 6.3.1 Exercises In Exercises 1 - 33, solve the equation analytically. 1. 24x = 8 4. 42x = 1 2 7. 37x = 814−2x 10. 5−x = 2 13. (1.005)12x = 3 16. 500 1 − e2x = 250 19. 100ex ex + 2 x 22. 25 4 5 = 50 = 10 25. 3(x−1) = 2x 2. 3(x−1) = 27 5. 8x = 1 128 8. 9 · 37x = 1 9 2x 11. 5x = −2 14. e−5730k = 1 2 3. 52x−1 = 125 6. 2(x3−x) = 1 9. 32x =
5 12. 3(x−1) = 29 15. 2000e0.1t = 4000 17. 70 + 90e−0.1t = 75 18. 30 − 6e−0.1x = 20 20. 5000 1 + 2e−3t = 2500 23. e2x = 2ex 26. 3(x−1) = 1 2 (x+5) 21. 150 1 + 29e−0.8t = 75 24. 7e2x = 28e−6x 27. 73+7x = 34−2x 28. e2x − 3ex − 10 = 0 29. e2x = ex + 6 30. 4x + 2x = 12 31. ex − 3e−x = 2 32. ex + 15e−x = 8 33. 3x + 25 · 3−x = 10 In Exercises 34 - 39, solve the inequality analytically. 34. ex > 53 36. 2(x3−x) < 1 38. 150 1 + 29e−0.8t ≤ 130 35. 1000 (1.005)12t ≥ 3000 37. 25 4 5 x ≥ 10 39. 70 + 90e−0.1t ≤ 75 In Exercises 40 - 45, use your calculator to help you solve the equation or inequality. 40. 2x = x2 41. ex = ln(x) + 5 43. e−x − xe−x ≥ 0 44. 3(x−1) < 2x √ 42. e x = x + 1 45. ex < x3 − x 46. Since f (x) = ln(x) is a strictly increasing function, if 0 < a < b then ln(a) < ln(b). Use this fact to solve the inequality e(3x−1) > 6 without a sign diagram. Use this technique to solve the inequalities in Exercises 34 - 39. (NOTE: Isolate the exponential function ﬁrst!) 47. Compute the inverse of f (x) = ex − e−x 2. State the domain and range of both f and f −1. 6.3 Exponential Equations and Inequalities 457 48. In Example 6.3.4, we found that the inverse of f (x) = we left a few loose ends for you to tie up. 5ex ex + 1 was f −
1(x) = ln x 5 − x but (a) Show that f −1 ◦ f (x) = x for all x in the domain of f and that f ◦ f −1 (x) = x for all x in the domain of f −1. (b) Find the range of f by ﬁnding the domain of f −1. 5x x + 1 (c) Let g(x) = and h(x) = ex. Show that f = g ◦ h and that (g ◦ h)−1 = h−1 ◦ g−1. (We know this is true in general by Exercise 31 in Section 5.2, but it’s nice to see a speciﬁc example of the property.) 49. With the help of your classmates, solve the inequality ex > xn for a variety of natural numbers n. What might you conjecture about the “speed” at which f (x) = ex grows versus any polynomial? 458 Exponential and Logarithmic Functions 6.3.2 Answers 1. x = 3 4 4. x = − 1 4 7. x = 16 15 10. x = − ln(2) ln(5) 13. x = ln(3) 12 ln(1.005) 2. x = 4 5. x = − 7 3 8. x = − 2 11 11. No solution. 14. k = ln( 1 2 ) −5730 = ln(2) 5730 3. x = 2 6. x = −1, 0, 1 9. x = ln(5) 2 ln(3) 12. x = ln(29)+ln(3) ln(3) 15. t = ln(2) 0.1 = 10 ln(2) 17. t = 18 ) ln( 1 −0.1 = 10 ln(18) 19. x = ln(2) 21. t = ln( 1 29 ) −0.8 = 5 4 ln(29) 23. x = ln(2) 25. x = ln(3) ln(3)−ln(2) 27. x = 4 ln(3)−3 ln(7) 7 ln(7)+2 ln(3) 16. x = 1 2
ln 1 2 = − 1 2 ln(2) 18. x = −10 ln 5 3 = 10 ln 3 5 20. t = 1 3 ln(2) 22. x = = ln(2)−ln(5) ln(4)−ln(5) 24. x = − 1 = 1 4 ln(2) ln( 2 5 ) ln( 4 5 ) 8 ln 1 ln(3)+5 ln( 1 2 ) ln(3)−ln( 1 2 ) 4 26. x = = ln(3)−5 ln(2) ln(3)+ln(2) 28. x = ln(5) 31. x = ln(3) 34. (ln(53), ∞) 36. (−∞, −1) ∪ (0, 1) 29. x = ln(3) 32. x = ln(3), ln(5) 35. 30. x = ln(3) ln(2) 33. x = ln(5) ln(3) −∞, ln(2)−ln(5) ln(4)−ln(5) = ln(3) 12 ln(1.005), ∞ 37. −∞, ln( 2 5 ) ln( 4 5 ) 38. −∞, ln( 2 377 ) −0.8 = −∞, 5 4 ln 377 2 39. ln( 1 18 ) −0.1, ∞ = [10 ln(18), ∞) 40. x ≈ −0.76666, x = 2, x = 4 41. x ≈ 0.01866, x ≈ 1.7115 42. x = 0 44. ≈ (−∞, 2.7095) 43. (−∞, 1] 45. ≈ (2.3217, 4.3717) 46. x > 1 3 (ln(6) + 1) √ 47. f −1 = ln x + x2 + 1. Both f and f −1 have domain (−∞, ∞) and range (−∞, ∞). 6.4 Logarithmic Equations and Inequalities 459 6.4 Logarithmic Equations and In
equalities In Section 6.3 we solved equations and inequalities involving exponential functions using one of two basic strategies. We now turn our attention to equations and inequalities involving logarithmic functions, and not surprisingly, there are two basic strategies to choose from. For example, suppose we wish to solve log2(x) = log2(5). Theorem 6.4 tells us that the only solution to this equation is x = 5. Now suppose we wish to solve log2(x) = 3. If we want to use Theorem 6.4, we need to 23 = log2(8). rewrite 3 as a logarithm base 2. We can use Theorem 6.3 to do just that: 3 = log2 Our equation then becomes log2(x) = log2(8) so that x = 8. However, we could have arrived at the same answer, in fewer steps, by using Theorem 6.3 to rewrite the equation log2(x) = 3 as 23 = x, or x = 8. We summarize the two common ways to solve log equations below. Steps for Solving an Equation involving Logarithmic Functions 1. Isolate the logarithmic function. 2. (a) If convenient, express both sides as logs with the same base and equate the arguments of the log functions. (b) Otherwise, rewrite the log equation as an exponential equation. Example 6.4.1. Solve the following equations. Check your solutions graphically using a calculator. 1. log117(1 − 3x) = log117 x2 − 3 2. 2 − ln(x − 3) = 1 3. log6(x + 4) + log6(3 − x) = 1 4. log7(1 − 2x) = 1 − log7(3 − x) 5. log2(x + 3) = log2(6 − x) + 3 6. 1 + 2 log4(x + 1) = 2 log2(x) Solution. 1. Since we have the same base on both sides of the equation log117(1 − 3x) = log117 x2 − 3, we equate what’s inside the logs to get 1 − 3x = x2 − 3. Solving x2 + 3x − 4 = 0 gives x = −4 and x = 1. To check these answers using the calculator, we make use of the change ln
(x2−3) of base formula and graph f (x) = ln(1−3x) ln(117) and we see they intersect only at x = −4. To see what happened to the solution x = 1, we substitute it into our original equation to obtain log117(−2) = log117(−2). While these expressions look identical, neither is a real number,1 which means x = 1 is not in the domain of the original equation, and is not a solution. ln(117) and g(x) = 2. Our ﬁrst objective in solving 2−ln(x−3) = 1 is to isolate the logarithm. We get ln(x−3) = 1, which, as an exponential equation, is e1 = x − 3. We get our solution x = e + 3. On the calculator, we see the graph of f (x) = 2 − ln(x − 3) intersects the graph of g(x) = 1 at x = e + 3 ≈ 5.718. 1They do, however, represent the same family of complex numbers. We stop ourselves at this point and refer the reader to a good course in Complex Variables. 460 Exponential and Logarithmic Functions y = f (x) = log117(1 − 3x) and x2 − 3 y = g(x) = log117 y = f (x) = 2 − ln(x − 3) and y = g(x) = 1 3. We can start solving log6(x + 4) + log6(3 − x) = 1 by using the Product Rule for logarithms to rewrite the equation as log6 [(x + 4)(3 − x)] = 1. Rewriting this as an exponential equation, we get 61 = (x + 4)(3 − x). This reduces to x2 + x − 6 = 0, which gives x = −3 and x = 2. Graphing y = f (x) = ln(x+4) and y = g(x) = 1, we see they intersect twice, at x = −3 and x = 2. ln(6) + ln(3−x) ln(6) y = f (x) = log6(x + 4) + log6(3 − x) and y
= g(x) = 1 4. Taking a cue from the previous problem, we begin solving log7(1 − 2x) = 1 − log7(3 − x) by ﬁrst collecting the logarithms on the same side, log7(1 − 2x) + log7(3 − x) = 1, and then using the Product Rule to get log7[(1 − 2x)(3 − x)] = 1. Rewriting this as an exponential equation gives 71 = (1−2x)(3−x) which gives the quadratic equation 2x2 −7x−4 = 0. Solving, we ﬁnd x = − 1 intersect ln(7) only at x = − 1 2. Checking x = 4 in the original equation produces log7(−7) = 1 − log7(−1), which is a clear domain violation. 2 and x = 4. Graphing, we ﬁnd y = f (x) = ln(1−2x) and y = g(x) = 1− ln(3−x) ln(7) 5. Starting with log2(x + 3) = log2(6 − x) + 3, we gather the logarithms to one side and get log2(x + 3) − log2(6 − x) = 3. We then use the Quotient Rule and convert to an exponential equation log2 x + 3 6 − x = 3 ⇐⇒ 23 = x + 3 6 − x This reduces to the linear equation 8(6 − x) = x + 3, which gives us x = 5. When we graph f (x) = ln(x+3) ln(2) + 3, we ﬁnd they intersect at x = 5. ln(2) and g(x) = ln(6−x) 6.4 Logarithmic Equations and Inequalities 461 y = f (x) = log7(1 − 2x) and y = g(x) = 1 − log7(3 − x) y = f (x) = log2(x + 3) and y = g(x) = log2(6 − x) + 3 6. Starting with 1 + 2 log4(x + 1) = 2 log2(x),
we gather the logs to one side to get the equation 1 = 2 log2(x) − 2 log4(x + 1). Before we can combine the logarithms, however, we need a common base. Since 4 is a power of 2, we use change of base to convert log4(x + 1) = log2(x + 1) log2(4) = 1 2 log2(x + 1) Hence, our original equation becomes 2 log2(x + 1) 1 = 2 log2(x) − 2 1 1 = 2 log2(x) − log2(x + 1) x2 − log2(x + 1) 1 = log2 x2 x + 1 1 = log2 Power Rule Quotient Rule Rewriting this in exponential form, we get x2 formula, we get x = 1 ± √ 3. Graphing f (x) = 1 + 2 ln(x+1) x+1 = 2 or x2 − 2x − 2 = 0. Using the quadratic ln(2), we see the 3 < 0, which means if and g(x) = 2 ln(x) ln(4) 3 ≈ 2.732. The solution x = 1 − √ graphs intersect only at x = 1 + substituted into the original equation, the term 2 log2 √ 3 is undeﬁned. 1 − √ y = f (x) = 1 + 2 log4(x + 1) and y = g(x) = 2 log2(x) 462 Exponential and Logarithmic Functions If nothing else, Example 6.4.1 demonstrates the importance of checking for extraneous solutions2 when solving equations involving logarithms. Even though we checked our answers graphically, extraneous solutions are easy to spot - any supposed solution which causes a negative number inside a logarithm needs to be discarded. As with the equations in Example 6.3.1, much can be learned from checking all of the answers in Example 6.4.1 analytically. We leave this to the reader and turn our attention to inequalities involving logarithmic functions. Since logarithmic functions are continuous on their domains, we can use sign diagrams. Example 6.4.2. Solve the following inequalities. Check your answer graphically using a calculator. 1. 1 ln(x
) + 1 ≤ 1 Solution. 2. (log2(x))2 < 2 log2(x) + 3 3. x log(x + 1) ≥ x 1. We start solving 1 1 1 ln(x) ln(x)+1 − ln(x)+1 ln(x)+1 ≤ 0 which reduces to − ln(x) ln(x)+1 ≤ 1 by getting 0 on one side of the inequality: Getting a common denominator yields ln(x)+1 ≥ 0. We deﬁne r(x) = ln(x) ln(x)+1 − 1 ≤ 0. ln(x)+1 ≤ 0, ln(x)+1 and set about ﬁnding the domain and the zeros or In order to keep the of r. Due to the appearance of the term ln(x), we require x > 0. denominator away from zero, we solve ln(x) + 1 = 0 so ln(x) = −1, so x = e−1 = 1 e. Hence, the domain of r is 0, 1 ln(x)+1 = 0 so that e ln(x) = 0, and we ﬁnd x = e0 = 1. In order to determine test values for r without resorting to the calculator, we need to ﬁnd numbers between 0, 1 e, and 1 which have a base of e. Since, we use the fact that e ≈ 2.718 > 1, 0 < 1 e < 1√ = −2 ln 1 −2+1 = 2, which is (+). The rest of the test values e2 ∪ [1, ∞). are determined similarly. From our sign diagram, we ﬁnd the solution to be 0, 1 e Graphing f (x) = 1 ln(x)+1 and g(x) = 1, we see the graph of f is below the graph of g on the solution intervals, and that the graphs intersect at x = 1. e2 < 1 = ln e−2 = −2, and ﬁnd r 1 e2 e, ∞. To ﬁnd the zeros of r, we set r(x) = ln(x) e < 1 < e. To determine the sign of r 1 ∪ 1 e2 (+
) (−) 0 (+) 0 1 e 1 2Recall that an extraneous solution is an answer obtained analytically which does not satisfy the original equation. y = f (x) = 1 ln(x)+1 and y = g(x) = 1 6.4 Logarithmic Equations and Inequalities 463 2. Moving all of the nonzero terms of (log2(x))2 < 2 log2(x) + 3 to one side of the inequality, we have (log2(x))2 − 2 log2(x) − 3 < 0. Deﬁning r(x) = (log2(x))2 − 2 log2(x) − 3, we get the domain of r is (0, ∞), due to the presence of the logarithm. To ﬁnd the zeros of r, we set r(x) = (log2(x))2 − 2 log2(x) − 3 = 0 which results in a ‘quadratic in disguise.’ We set u = log2(x) so our equation becomes u2 − 2u − 3 = 0 which gives us u = −1 and u = 3. Since u = log2(x), we get log2(x) = −1, which gives us x = 2−1 = 1 2, and log2(x) = 3, which yields x = 23 = 8. We use test values which are powers of 2: 0 < 1 2 < 1 < 8 < 16, and from our 2 sign diagram, we see r(x) < 0 on 1 is below the graph of y = g(x) = 2 ln(x) 2, 8. Geometrically, we see the graph of f (x) = ln(2) + 3 on the solution interval. ln(x) ln(2) 4 < 1 (+) 0 (−) 1 2 0 (+) 8 0 y = f (x) = (log2(x))2 and y = g(x) = 2 log2(x) + 3 3. We begin to solve x log(x+1) ≥ x by subtracting x from both sides to get x log(x+1)−x ≥ 0. We deﬁne r(x) = x log(x+1)−x and due to the presence
of the logarithm, we require x+1 > 0, or x > −1. To ﬁnd the zeros of r, we set r(x) = x log(x + 1) − x = 0. Factoring, we get x (log(x + 1) − 1) = 0, which gives x = 0 or log(x+1)−1 = 0. The latter gives log(x+1) = 1, or x + 1 = 101, which admits x = 9. We select test values x so that x + 1 is a power of 10, and we obtain −1 < −0.9 < 0 < 10 − 1 < 9 < 99. Our sign diagram gives the solution to be (−1, 0] ∪ [9, ∞). The calculator indicates the graph of y = f (x) = x log(x + 1) is above y = g(x) = x on the solution intervals, and the graphs intersect at x = 0 and x = 9. √ (+) 0 (−) 0 (+) −1 0 9 y = f (x) = x log(x + 1) and y = g(x) = x 464 Exponential and Logarithmic Functions Our next example revisits the concept of pH ﬁrst seen in Exercise 77 in Section 6.1. Example 6.4.3. In order to successfully breed Ippizuti ﬁsh the pH of a freshwater tank must be at least 7.8 but can be no more than 8.5. Determine the corresponding range of hydrogen ion concentration, and check your answer using a calculator. Solution. Recall from Exercise 77 in Section 6.1 that pH = − log[H+] where [H+] is the hydrogen ion concentration in moles per liter. We require 7.8 ≤ − log[H+] ≤ 8.5 or −7.8 ≥ log[H+] ≥ −8.5. To solve this compound inequality we solve −7.8 ≥ log[H+] and log[H+] ≥ −8.5 and take the intersection of the solution sets.3 The former inequality yields 0 < [H+] ≤ 10−7.8 and the latter yields [H+] ≥ 10−8.5. Taking the intersection gives us our ﬁnal answer 10−8.5 ≤ [H+] ≤
10−7.8. (Your Chemistry professor may want the answer written as 3.16 × 10−9 ≤ [H+] ≤ 1.58 × 10−8.) After carefully adjusting the viewing window on the graphing calculator we see that the graph of f (x) = − log(x) lies between the lines y = 7.8 and y = 8.5 on the interval [3.16 × 10−9, 1.58 × 10−8]. The graphs of y = f (x) = − log(x), y = 7.8 and y = 8.5 We close this section by ﬁnding an inverse of a one-to-one function which involves logarithms. Example 6.4.4. The function f (x) = log(x) 1 − log(x) is one-to-one. Find a formula for f −1(x) and check your answer graphically using your calculator. Solution. We ﬁrst write y = f (x) then interchange the x and y and solve for y. y = f (x) y = x = log(x) 1 − log(x) log(y) 1 − log(y) x (1 − log(y)) = log(y) x − x log(y) = log(y) Interchange x and y. x = x log(y) + log(y) x = (x + 1) log(y) x x + 1 = log(y) y = 10 x x+1 Rewrite as an exponential equation. 3Refer to page 4 for a discussion of what this means. 6.4 Logarithmic Equations and Inequalities 465 We have f −1(x) = 10 x x+1. Graphing f and f −1 on the same viewing window yields y = f (x) = log(x) 1 − log(x) and y = g(x) = 10 x x+1 466 Exponential and Logarithmic Functions 6.4.1 Exercises In Exercises 1 - 24, solve the equation analytically. 1. log(3x − 1) = log(4 − x) 3. ln 8 − x2 = ln(2 − x) 5. log3(7 − 2x) = 2 7. ln x2 − 99 = 0 9.

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