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Example 3.12. Let \( f \) be the function given by the following graph.
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Solution. Most of the answers can be read immediately from the graph.\n\n(a) For the domain, we project the graph to the \( x \) -axis. The domain consists of all numbers from -5 to 5 without -3, that is \( D = \left\lbrack {-5, - 3)\cup ( - 3,5}\right\rbrack \) .\n\n(b) For the range, we project the graph to the \( y \) -axis. The domain consists of all numbers from 1 to 5, that is \( R = \left\lbrack {1,5}\right\rbrack \) .\n\n(c) To find \( x \) with \( f\left( x\right) = 3 \) we look at the horizontal line at \( y = 3 \) :\n\nWe see that there are two numbers \( x \) with \( f\left( x\right) = 3 \) :\n\n\[ f\left( {-2}\right) = 3,\;f\left( 3\right) = 3. \]\n\nTherefore, the answer is \( x = - 2 \) or \( x = 3 \) .\n\n(d) Looking at the horizontal line \( y = 2 \), we see that there is only one \( x \) with \( f\left( x\right) = 2 \) ; namely \( f\left( 4\right) = 2 \) . Note, that \( x = - 3 \) does not solve the problem, since \( f\left( {-3}\right) \) is undefined. The answer is \( x = 4 \) .\n\n(e) To find \( x \) with \( f\left( x\right) > 2 \), the graph has to lie above the line \( y = 2 \) .\n\nWe see that the answer is those numbers \( x \) greater than -3 and less than 4. The answer is therefore the interval \( \left( {-3,4}\right) \) .\n\n(f) For \( f\left( x\right) \leq 4 \), we obtain all numbers \( x \) from the domain that are less than -1 or greater or equal to 1 . The answer is therefore,\n\n\[ \lbrack - 5, - 3) \cup \left( {-3, - 1}\right) \cup \left\lbrack {1,5}\right\rbrack \text{.} \]\n\nNote that -3 is not part of the answer, since \( f\left( {-3}\right) \) is undefined.\n\n(g) \( f\left( 1\right) = 4 \), and \( f\left( 4\right) = 2 \)\n\n(h) \( f\left( 1\right) + f\left( 4\right) = 4 + 2 = 6 \)\n\n(i) \( f\left( 1\right) + 4 = 4 + 4 = 8 \)\n\n(j) \( f\left( {1 + 4}\right) = f\left( 5\right) = 1 \)
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Yes
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a) What was the population size in the years 2004 and 2009?\nb) In what years did the city have the most population?\nc) In what year did the population grow the fastest?\nd) In what year did the population decline the fastest?
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Solution. The population size in the year 2004 was approximately 36,000 . In the year 2009, it was approximately 26,000 . The largest population was in the year 2006, where the graph has its maximum. The fastest growth in the population was between the years 2003 and 2004. That is when the graph has the largest slope. Finally, the fastest decline happened in the years 2006-2007.
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Yes
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Graph the function described by the following formula:\n\n\\[ f\\left( x\\right) = \\left\\{ \\begin{matrix} x + 3 & ,\\text{ for } & - 3 \\leq x < - 1 \\\\ {x}^{2} & ,\\text{ for } & - 1 < x < 1 \\\\ 3 & ,\\text{ for } & 2 < x \\leq 3 \\end{matrix}\\right. \\]
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Solution. We really have to graph all three functions \\( y = x + 3 \\) , \\( y = {x}^{2} \\), and \\( y = 3 \\), and then restrict them to their respective domain. Graphing the three functions, we obtain the following tables and associated graphs, which we draw in one \\( x - y \\) -plane:\n\n<table><thead><tr><th colspan=\
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No
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Graph the equations \( y = \sqrt{7 - x} \) and \( y = {x}^{3} - 2{x}^{2} - 4 \) in the same window.
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Solution. Enter the functions as Y1 and Y2 after pressing \( \left( \overline{y = }\right) \) . The graphs of both functions appear together in the graphing window:
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No
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Graph the relation \( {\left( x - 3\right) }^{2} + {\left( y - 5\right) }^{2} = {16} \) .
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Since the above expression is not solved for \( y \), we cannot simply plug this into the calculator. Instead, we have to solve for \( y \) first.\n\n\[{\left( x - 3\right) }^{2} + {\left( y - 5\right) }^{2} = {16}\; \Rightarrow \;{\left( y - 5\right) }^{2} = {16} - {\left( x - 3\right) }^{2}\]\n\n\[ \Rightarrow \;y - 5 = \pm \sqrt{{16} - {\left( x - 3\right) }^{2}}\]\n\n\[ \Rightarrow \;y = 5 \pm \sqrt{{16} - {\left( x - 3\right) }^{2}}\]\n\nNote, that there are two functions that we need to graph: \( y = 5 + \sqrt{{16} - {\left( x - 3\right) }^{2}} \) and \( y = 5 - \sqrt{{16} - {\left( x - 3\right) }^{2}} \) . Entering these as Y1 and Y2 in the calculator gives the following function menu and graphing window:\n\n\n\nThe graph displays a circle of radius 4 with a center at \( \left( {3,5}\right) \) . However, due to the current scaling, the graph looks more like an ellipse instead of a circle. We can remedy this by using the \
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Yes
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Example 4.6. Graph the equation \( y = {x}^{3} - 2{x}^{2} - {4x} + 4 \).
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Solution.\n\na) Graphing the function in the standard window gives the following graph:\n\n\n\n, we can move a cursor to the right and left via We can see several function values as displayed \n\nFor the graph on the left we have displayed the point \( \left( {x, y}\right) \approx \) \( \left( {-{.638},{5.478}}\right) \), and for the graph on the right, we have displayed the point \( \left( {x, y}\right) \approx \left( {{2.128}, - {3.933}}\right) \) .\n\nAnother way to find function values is to look at the table menu by pressing 2nd graph graph :\n\n\n\nWe see the \( y \) -values for specific \( x \) -values are displayed. More values can be displayed by pressing \( \left( \Delta \right) \) and \( \left( \nabla \right) \) . Furthermore, the settings for the table can be changed in the table-set menu by pressing 2nd (window). In particular, by changing the the \
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Yes
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Graph the equation \( y = {x}^{4} - {x}^{3} - 4{x}^{2} + {4x} \) .
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Solution.\na) First, press \( \left( \overline{\;y = \;}\right) \) and enter the function \( y = {x}^{4} - {x}^{3} - 4{x}^{2} + {4x} \) . We find the zeros using item 2 from the calculate menu. Here are two of the four zeros:\n\n\n\nNote, in particular, the left zero has a \( y \) value of \
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No
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Solve the equation. Approximate your answer to the nearest hundredth.\n\n\[ \n\text{a)}{x}^{4} + 3{x}^{2} - {5x} - 7 = 0\text{, b)}{x}^{3} - 4 = {7x} - {3}^{x}\text{.} \n\]
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Solution. a) We need to find all numbers \( x \) so that \( {x}^{4} + 3{x}^{2} - {5x} - 7 = 0 \) . Note, that these are precisely the zeros of the function \( f\left( x\right) = {x}^{4} + 3{x}^{2} - {5x} - 7 \) , since the zeros are the values \( x \) for which \( f\left( x\right) = 0 \) . Graphing the function \( f\left( x\right) = {x}^{4} + 3{x}^{2} - {5x} - 7 \) shows that there are two zeros.\n\n\n\nUsing the calculate menu, we can identify both zeros.\n\n\n\nThe answer is therefore, \( x \approx - {0.85} \) and \( x \approx {1.64} \) (rounded to the nearest hundredth).
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No
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Graph the equations \( y = {x}^{2} - {3x} + 2 \) and \( y = {x}^{3} + 2{x}^{2} - 1 \).
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Solution. First, enter the two equations for Y1 and Y2 after pressing the y= ) key. Both graphs are displayed in the graphing window.\n\n\n\na) The procedure for finding the intersection of the graphs for Y1 and Y2 is similar to finding minima, maxima, or zeros. In the calculate menu, 2nd ) trace intersect (item 5). Choose the first curve Y1 (with \( \bigtriangleup \) , \( \nabla \), and enter ), and the second curve\n\nY2. Finally choose a guess of the intersection point (with \( \vartriangleleft \) ,\n\n\( \left( \overset{\overleftrightarrow{} }{ \vartriangleright }\right) \), and enter enter the intersection is approximated as \( \left( {x, y}\right) \approx \)\n\n\n\nb) The intersection is the point where the two \( y \) -values \( y = {x}^{2} - {3x} + 2 \) and \( y = {x}^{3} + 2{x}^{2} - 1 \) coincide. Therefore, we want to find an \( x \) with\n\n\[ \n{x}^{2} - {3x} + 2 = {x}^{3} + 2{x}^{2} - 1 \n\]\n\nThe TI-84 equation solver requires an equation with zero on the left. Therefore, we subtract the left-hand side, and obtain:\n\n\[ \n\text{ (subtract }\left( {{x}^{2} - {3x} + 2}\right) )\; \Rightarrow \;0 = \left( {{x}^{3} + 2{x}^{2} - 1}\right) - \left( {{x}^{2} - {3x} + 2}\right) \n\]\n\n\[ \n\Rightarrow \;0 = {x}^{3} + 2{x}^{2} - 1 - {x}^{2} + {3x} - 2 \n\]\n\n\[ \n\Rightarrow \;0 = {x}^{3} + {x}^{2} + {3x} - 3 \n\]\n\nEntering \( 0 = {x}^{3} + {x}^{2} + {3x} - 3 \) into the equation solver, \( \left( \text{ math }\right) \left( \Delta \right) \) enter \( ) \), and choosing an approximation \( x = 1 \) for \( \mathrm{X} \), the calculator solves this equation (using alpha enter enter ) as follows: \n\nThe approximation provided by the calculator is \( x \approx .{71134573927}\ldots \) .
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Yes
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Example 5.6. Guess the formula for the function, based on the basic graphs in Section 5.1 and the transformations described above.
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Solution. a) This is the square-root function shifted to the left by 2 . Thus, by Observation 5.1, this is the function \( f\left( x\right) = \sqrt{x + 2} \) .\nb) This is the graph of \( y = \frac{1}{x} \) reflected about the \( x \) -axis (or also \( y = \frac{1}{x} \) reflected about the \( y \) -axis). In either case, we obtain the rule \( y = - \frac{1}{x} \) .\nc) This is a parabola reflected about the \( x \) -axis and then shifted up by 3 . Thus, we get:\n\n\[ y = {x}^{2} \]\n\nreflecting about the \( x \) -axis gives \( \;y = - {x}^{2} \)\n\n\[ \text{shifting the graph up by 3 gives}y = - {x}^{2} + 3 \]\n\nd) Starting from the graph of the cubic equation \( y = {x}^{3} \), we need to reflect about the \( x \) -axis (or also \( y \) -axis), then shift up by 2 and to the right by 3 . These transformations affect the formula as follows:\n\n\[ y = {x}^{3} \]\n\nreflecting about the \( x \) -axis gives \( \;y = - {x}^{3} \)\n\nshifting up by 2 gives \( \;y = - {x}^{3} + 2 \)\n\nshifting the the right by 3 gives \( \;y = - {\left( x - 3\right) }^{3} + 2 \)
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Yes
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Sketch the graph of the function, based on the basic graphs in Section 5.1 and the transformations described above.
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Solution. a) This is the parabola \( y = {x}^{2} \) shifted up by 3 . The graph is shown below.\nb) \( y = {\left( x + 2\right) }^{2} \) is the parabola \( y = {x}^{2} \) shifted 2 units to the left.  \nc) The graph of the function \( f\left( x\right) = \left| {x - 3}\right| - 2 \) is the absolute value shifted to the right by 3 and down by 2 . (Alternatively, we can first shift down by 2 and then to the right by 3.)\nd) Similarly, to get from the graph of \( y = \sqrt{x} \) to the graph of \( y = \sqrt{x + 1} \) , we shift the graph to the left, and then for \( y = 2 \cdot \sqrt{x + 1} \), we need to stretch the graph by a factor 2 away from the \( x \) -axis. (Alternatively, we could first stretch the the graph away from the \( x \) -axis, then shift the graph by 1 to the left.) \ne) For \( y = - \left( {\frac{1}{x} + 2}\right) \), we start with \( y = \frac{1}{x} \) and add 2, giving \( y = \frac{1}{x} + 2 \), which shifts the graph up by 2 . Then, taking the negative gives \( y = - \left( {\frac{1}{x} + 2}\right) \) , which corresponds to reflecting the graph about the \( x \) -axis.\nNote, that in this case, we cannot perform these transformations in the opposite order, since the negative of \( y = \frac{1}{x} \) gives \( y = - \frac{1}{x} \), and adding 2 gives \( y = - \frac{1}{x} + 2 \) which is not equal to \( - \left( {\frac{1}{x} + 2}\right) \) .\nf) We start with \( y = {x}^{3} \) . Adding 1 in the argument, \( y = {\left( x + 1\right) }^{3} \), shifts its graph to the left by 1 . Then, taking the negative in the argument gives \( y = {\left( -x + 1\right) }^{3} \), which reflects the graph about the \( y \) -axis.\nHere, the order in which we perform these transformations is again important. In fact, if we first take the negative in the argument, we obtain \( y = {\left( -x\right) }^{3} \) . Then, adding one in the argument would give \( y = {\left( -\left( x + 1\right) \right) }^{3} = \) \( {\left( -x - 1\right) }^{3} \) which is different than our given function \( y = {\left( -x + 1\right) }^{3} \) .\n\nAll these solutions may also easily be checked by using the graphing function of the calculator.
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Yes
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a) The graph of \( f\left( x\right) = \left| {{x}^{3} - 5}\right| \) is stretched away from the \( y \) -axis by a factor of 3 . What is the formula for the new function?
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a) By Observation 5.4 on page 67 we have to multiply the argument by \( \frac{1}{3} \) . The new function is therefore:\n\n\[ f\left( {\frac{1}{3} \cdot x}\right) = \left| {{\left( \frac{1}{3} \cdot x\right) }^{3} - 5}\right| = \left| {\frac{1}{27} \cdot {x}^{3} - 5}\right| \]
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Yes
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Example 5.10. Determine, if the following functions are even, odd, or neither.
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Solution. The function \( f\left( x\right) = {x}^{2} \) is even, since \( f\left( {-x}\right) = {\left( -x\right) }^{2} = {x}^{2} \) . Similarly, \( g\left( x\right) = {x}^{3} \) is odd, \( h\left( x\right) = {x}^{4} \) is even, and \( k\left( x\right) = {x}^{5} \) is odd, since\n\n\[ g\left( {-x}\right) = {\left( -x\right) }^{3} = - {x}^{3} = - g\left( x\right) \]\n\n\[ h\left( {-x}\right) = {\left( -x\right) }^{4} = {x}^{4} = h\left( x\right) \]\n\n\[ k\left( {-x}\right) = {\left( -x\right) }^{5} = - {x}^{5} = - k\left( x\right) \]\n\nIndeed, we see that a function \( y = {x}^{n} \) is even, precisely when \( n \) is even, and \( y = {x}^{n} \) is odd, precisely when \( n \) is odd. (These examples are in fact the motivation behind defining even and odd functions as in definition 5.9 above.)
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Yes
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Example 6.1. Let \( f\left( x\right) = {x}^{2} + {5x} \) and \( g\left( x\right) = {7x} - 3 \) . Find the following functions, and state their domain.\n\n\[ \left( {f + g}\right) \left( x\right) ,\left( {f - g}\right) \left( x\right) ,\left( {f \cdot g}\right) \left( x\right) ,\text{ and }\left( \frac{f}{g}\right) \left( x\right) . \]
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Solution. The functions are calculated by adding the functions (or subtracting, multiplying, dividing).\n\n\[ \left( {f + g}\right) \left( x\right) = \left( {{x}^{2} + {5x}}\right) + \left( {{7x} - 3}\right) = {x}^{2} + {12x} - 3, \]\n\n\[ \left( {f - g}\right) \left( x\right) = \left( {{x}^{2} + {5x}}\right) - \left( {{7x} - 3}\right) \]\n\n\[ = {x}^{2} + {5x} - {7x} + 3 = {x}^{2} - {2x} + 3 \]\n\n\[ \left( {f \cdot g}\right) \left( x\right) = \left( {{x}^{2} + {5x}}\right) \cdot \left( {{7x} - 3}\right) \]\n\n\[ = 7{x}^{3} - 3{x}^{2} + {35}{x}^{2} - {15x} = 7{x}^{3} + {32}{x}^{2} - {15x}, \]\n\n\[ \left( \frac{f}{g}\right) \left( x\right) = \frac{{x}^{2} + {5x}}{{7x} - 3}. \]\n\nThe calculation of these functions was straightforward. To state their domain is also straightforward, except for the domain of the quotient \( \frac{f}{g} \) . Note, that \( f + g, f - g \), and \( f \cdot g \) are all polynomials. By the standard convention (Convention 3.4 on page [36) all these functions have the domain \( \mathbb{R} \), that is their domain is all real numbers.\n\nNow, for the domain of \( \frac{f}{g} \), we have to be a bit more careful, since the denominator of a fraction cannot be zero. The denominator of \( \frac{f}{g}\left( x\right) = \frac{{x}^{2} + {5x}}{{7x} - 3} \) is zero, exactly when\n\n\[ {7x} - 3 = 0\; \Rightarrow \;{7x} = 3\; \Rightarrow \;x = \frac{3}{7}. \]\n\nWe have to exclude \( \frac{3}{7} \) from the domain. The domain of the quotient \( \frac{f}{g} \) is therefore \( \mathbb{R} - \left\{ \frac{3}{7}\right\} \) .
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Yes
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Example 6.3. Let \( f\left( x\right) = \sqrt{x + 2} \), and let \( g\left( x\right) = {x}^{2} - {5x} + 4 \) . Find the functions \( \frac{f}{g} \) and \( \frac{g}{f} \) and state their domains.
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Solution. First, the domain of \( f \) consists of those numbers \( x \) for which the square root is defined. In other words, we need \( x + 2 \geq 0 \), that is \( x \geq - 2 \), so that the domain of \( f \) is \( {D}_{f} = \lbrack - 2,\infty ) \) . On the other hand, the domain of \( g \) is all real numbers, \( {D}_{g} = \mathbb{R} \) . Now, we have the quotients\n\n\[ \left( \frac{f}{g}\right) \left( x\right) = \frac{\sqrt{x + 2}}{{x}^{2} - {5x} + 4},\;\text{ and }\;\left( \frac{g}{f}\right) \left( x\right) = \frac{{x}^{2} - {5x} + 4}{\sqrt{x + 2}}. \]\n\nFor the domain of \( \frac{f}{g} \), we need to exclude those numbers \( x \) for which \( {x}^{2} - \) \( {5x} + 4 = 0 \) . Thus,\n\n\[ {x}^{2} - {5x} + 4 = 0 \Rightarrow \;\left( {x - 1}\right) \left( {x - 4}\right) = 0 \]\n\n\[ \Rightarrow x = 1\text{, or}x = 4\text{.} \]\n\nWe obtain the domain for \( \frac{f}{g} \) as the combined domain for \( f \) and \( g \), and exclude 1 and 4. Therefore, \( {D}_{\underline{f}} = \lbrack - 2,\infty ) - \{ 1,4\} \) .\n\nNow, for \( \frac{g}{f}\left( x\right) = \frac{{x}^{2} - {5x} + 4}{\sqrt{x + 2}} \), the denominator becomes zero exactly when\n\n\[ x + 2 = 0,\; \Rightarrow \;x = - 2. \]\n\nTherefore, we need to exclude -2 from the domain, that is\n\n\[ {D}_{\frac{g}{f}} = \lbrack - 2,\infty ) - \{ - 2\} = \left( {-2,\infty }\right) . \]
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Yes
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To form the quotient \( \frac{f}{g}\left( x\right) \) where \( f\left( x\right) = {x}^{2} - 1 \) and \( g\left( x\right) = x + 1 \)
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we write \( \frac{f}{g}\left( x\right) = \frac{{x}^{2} - 1}{x + 1} = \frac{\left( {x + 1}\right) \left( {x - 1}\right) }{x + 1} = x - 1 \) . One might be tempted to say that the domain is all real numbers. But it is not! The domain is all real numbers except -1 , and the last step of the simplification performed above is only valid for \( x \neq - 1 \) .
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Yes
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Example 6.6. Let \( f\left( x\right) = 2{x}^{2} + {5x} \) and \( g\left( x\right) = 2 - x \) . Find the following compositions\n\n\[ \n\\text{a)}f\\left( {g\\left( 3\\right) }\\right) ,\\;\\text{b)}g\\left( {f\\left( 3\\right) }\\right) ,\\;\\text{c)}f\\left( {f\\left( 1\\right) }\\right) ,\\;\\text{d)}f\\left( {2 \\cdot g\\left( 5\\right) }\\right) ,\\;\\text{e)}g\\left( {g\\left( 4\\right) + 5}\\right) \\text{.} \n\]
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Solution. We evaluate the expressions, one at a time, as follows:\n\n\[ \n\\text{a)}\\;f\\left( {g\\left( 3\\right) }\\right) = f\\left( {2 - 3}\\right) = f\\left( {-1}\\right) = 2 \\cdot {\\left( -1\\right) }^{2} + 5 \\cdot \\left( {-1}\\right) \n\]\n\n\[ \n= 2 - 5 = - 3 \n\]\n\nb) \\;g\\left( {f\\left( 3\\right) }\\right) = g\\left( {2 \\cdot {3}^{2} + 5 \\cdot 3}\\right) = g\\left( {{18} + {15}}\\right) = g\\left( {33}\\right) \n\n\[ \n= 2 - {33} = - {31} \n\]\n\nc) \\;f\\left( {f\\left( 1\\right) }\\right) = f\\left( {2 \\cdot {1}^{2} + 5 \\cdot 1}\\right) = f\\left( {2 + 5}\\right) = f\\left( 7\\right) \n\n\[ \n= 2 \\cdot {7}^{2} + 5 \\cdot 7 = {98} + {35} = {133}, \n\]\n\nd) \\;f\\left( {2 \\cdot g\\left( 5\\right) }\\right) = f\\left( {2 \\cdot \\left( {2 - 5}\\right) }\\right) = f\\left( {2 \\cdot \\left( {-3}\\right) }\\right) = f\\left( {-6}\\right) \n\n\[ \n= 2 \\cdot {\\left( -6\\right) }^{2} + 5 \\cdot \\left( {-6}\\right) = {72} - {30} = {42}, \n\]\n\n\[ \n\\text{e)}g\\left( {g\\left( 4\\right) + 5}\\right) = g\\left( {\\left( {2 - 4}\\right) + 5}\\right) = g\\left( {\\left( {-2}\\right) + 5}\\right) = g\\left( 3\\right) = 2 - 3 = - 1\\text{.} \n\]
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Yes
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Find \( \left( {f \circ g}\right) \left( x\right) \) and \( \left( {g \circ f}\right) \left( x\right) \) for the following functions, and state their domains.
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a) Composing \( f \circ g \), we obtain\n\n\[ \left( {f \circ g}\right) \left( x\right) = f\left( {g\left( x\right) }\right) = \frac{3}{g\left( x\right) + 2} = \frac{3}{{x}^{2} - {3x} + 2}. \]\n\nThe domain is the set of numbers \( x \) for which the denominator is non-zero.\n\n\[ {x}^{2} - {3x} + 2 = 0\; \Rightarrow \;\left( {x - 2}\right) \left( {x - 1}\right) = 0 \]\n\n\[ \Rightarrow x = 2\text{ or }x = 1 \]\n\n\[ \Rightarrow \;{D}_{f \circ g} = \mathbb{R} - \{ 1,2\} \text{.} \]\n\nSimilarly,\n\n\[ \left( {g \circ f}\right) \left( x\right) = g\left( {f\left( x\right) }\right) = f{\left( x\right) }^{2} - {3f}\left( x\right) = {\left( \frac{3}{x + 2}\right) }^{2} - 3\frac{3}{x + 2} \]\n\n\[ = \frac{9}{{\left( x + 2\right) }^{2}} - \frac{9}{x + 2} = \frac{9 - 9\left( {x + 2}\right) }{{\left( x + 2\right) }^{2}} \]\n\n\[ = \frac{9 - {9x} - {18}}{{\left( x + 2\right) }^{2}} = \frac{-{9x} - 9}{{\left( x + 2\right) }^{2}} = \frac{-9 \cdot \left( {x + 1}\right) }{{\left( x + 2\right) }^{2}} \]\n\nThe domain is all real numbers except \( x = - 2 \), that is \( {D}_{g \circ f} = \mathbb{R} - \{ - 2\} \) .
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Yes
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Example 6.9. Let \( f \) and \( g \) be the functions defined by the following table.\n\n<table><thead><tr><th>\( x \)</th><th>1</th><th>2</th><th>3</th><th>4</th><th>5</th><th>6</th><th>7</th></tr></thead><tr><td>\( f\left( x\right) \)</td><td>6</td><td>3</td><td>1</td><td>4</td><td>0</td><td>7</td><td>6</td></tr><tr><td>\( g\left( x\right) \)</td><td>4</td><td>0</td><td>2</td><td>5</td><td>\( - 2 \)</td><td>3</td><td>1</td></tr></table>\n\nDescribe the following functions via a table:\n\n\[ \n\\text{a)}2 \\cdot f\\left( x\\right) + 3,\\;\\text{b)}f\\left( x\\right) - g\\left( x\\right) ,\\;\\text{c)}f\\left( {x + 2}\\right) ,\\;\\text{d)}g\\left( {-x}\\right) \\text{.} \n\]
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Solution. For (a) and (b), we obtain by immediate calculation\n\n<table><thead><tr><th>\( x \)</th><th>1</th><th>2</th><th>3</th><th>4</th><th>5</th><th>6</th><th>7</th></tr></thead><tr><td>\( 2 \cdot f\left( x\right) + 3 \)</td><td>15</td><td>9</td><td>5</td><td>11</td><td>3</td><td>17</td><td>15</td></tr><tr><td>\( f\left( x\right) - g\left( x\right) \)</td><td>2</td><td>3</td><td>\( - 1 \)</td><td>\( - 1 \)</td><td>2</td><td>4</td><td>5</td></tr></table>\n\nFor example, for \( x = 3 \), we obtain \( 2 \cdot f\left( x\right) + 3 = 2 \cdot f\left( 3\right) + 3 = 2 \cdot 1 + 3 = 5 \) and \( f\left( x\right) - g\left( x\right) = f\left( 3\right) - g\left( 3\right) = 1 - 2 = - 1 \) .
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No
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Example 6.10. Let \( f \) and \( g \) be the functions defined by the following table.\n\n<table><thead><tr><th>\( x \)</th><th>1</th><th>3</th><th>5</th><th>7</th><th>9</th><th>11</th></tr></thead><tr><td>\( f\left( x\right) \)</td><td>3</td><td>5</td><td>11</td><td>4</td><td>9</td><td>7</td></tr><tr><td>\( g\left( x\right) \)</td><td>7</td><td>\( - 6 \)</td><td>9</td><td>11</td><td>9</td><td>5</td></tr></table>\n\nDescribe the following functions via a table:\n\n\[ \text{a)}f \circ g,\;\text{b)}g \circ f,\;\text{c)}f \circ f,\;\text{d)}g \circ g\text{.} \]
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Solution. The compositions are calculated by repeated evaluation. For example,\n\n\[ \left( {f \circ g}\right) \left( 1\right) = f\left( {g\left( 1\right) }\right) = f\left( 7\right) = 4. \]\n\nThe complete answer is displayed below.\n\n<table><thead><tr><th>\( x \)</th><th>1</th><th>3</th><th>5</th><th>7</th><th>9</th><th>11</th></tr></thead><tr><td>\( \left( {f \circ g}\right) \left( x\right) \)</td><td>4</td><td>undef.</td><td>9</td><td>7</td><td>9</td><td>11</td></tr><tr><td>\( \left( {g \circ f}\right) \left( x\right) \)</td><td>\( - 6 \)</td><td>9</td><td>5</td><td>undef.</td><td>9</td><td>11</td></tr><tr><td>\( \left( {f \circ f}\right) \left( x\right) \)</td><td>5</td><td>11</td><td>7</td><td>undef.</td><td>9</td><td>4</td></tr><tr><td>\( \left( {g \circ g}\right) \left( x\right) \)</td><td>11</td><td>undef.</td><td>9</td><td>5</td><td>9</td><td>9</td></tr></table>
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Yes
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As was noted above, the function \( f\left( x\right) = {x}^{2} \) is not one-to-one, because, for example, for inputs 2 and -2 , we have the same output
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\[ f\left( {-2}\right) = {\left( -2\right) }^{2} = 4,\;f\left( 2\right) = {2}^{2} = 4. \]
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Yes
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c) \( f\left( x\right) = - {x}^{3} + 6{x}^{2} - {13x} + {12} \)
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Solution. We use the horizontal line test to see which functions are one-toone. For (a) and (b), we see that the functions are not one-to-one since there\nis a horizontal line that intersects with the graph more than once: \n\nFor (c), using the calculator to graph the function \( f\left( x\right) = - {x}^{3} + 6{x}^{2} - {13x} + {12} \) , we see that all horizontal lines intersect the graph exactly once. Therefore the function in part (c) is one-to-one. The function in part (d) however has a graph that intersects some horizontal line in several points. Therefore \( f\left( x\right) = {x}^{3} - 2{x}^{2} + 3 \) is not one-to-one:
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Yes
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Find the inverse of the following function via algebra.\na) \( f\left( x\right) = {2x} - 7 \)
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a) First, reverse the role of input and output in \( y = {2x} - 7 \) by exchanging the variables \( x \) and \( y \) . That is, we write \( x = {2y} - 7 \) . We need to solve this for \( y \) :\n\n\[ \overset{\left( \text{ add }7\right) }{ \Rightarrow }\;x + 7 = {2y} \Rightarrow \;y = \frac{x + 7}{2}. \]\n\nTherefore, we obtain that the inverse of \( f \) is \( {f}^{-1}\left( x\right) = \frac{x + 7}{2} \).
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Yes
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Note, that the function \( y = {x}^{2} \) can be restricted to a one-to-one function by choosing the domain to be all non-negative numbers \( \lbrack 0,\infty ) \) or by choosing the domain to be all non-positive numbers \( ( - \infty ,0\rbrack \) .
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Let \( f : \lbrack 0,\infty ) \rightarrow \lbrack 0,\infty ) \) be the function \( f\left( x\right) = {x}^{2} \), so that \( f \) has a domain of all non-negative numbers. Then, the inverse is the function \( {f}^{-1}\left( x\right) = \sqrt{x} \). On the other hand, we can take \( g\left( x\right) = {x}^{2} \) whose domain consists of all non-positive numbers \( ( - \infty ,0\rbrack \), that is \( g : \left( {-\infty ,0\rbrack \rightarrow \lbrack 0,\infty }\right) \). Then, the inverse function \( {g}^{-1} \) must reverse domain and range, that is \( {g}^{-1} : \lbrack 0,\infty ) \rightarrow ( - \infty ,0\rbrack \). The inverse is obtained by exchanging \( x \) and \( y \) in \( y = {x}^{2} \) as follows:\n\n\[ x = {y}^{2}\; \Rightarrow y = \pm \sqrt{x}\; \Rightarrow {g}^{-1}\left( x\right) = - \sqrt{x}. \]
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Yes
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Example 7.8. Restrict the function to a one-to-one function. Find the inverse function, if possible.
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Solution. The graphs of \( f \) and \( g \) are displayed below.\na)\n\na) The graph shows that \( f \) is one-to-one when restricted to all numbers \( x \geq - 3 \), which is the choice we make to find an inverse function. Next, we replace \( x \) and \( y \) in \( y = {\left( x + 3\right) }^{2} + 1 \) to give \( x = {\left( y + 3\right) }^{2} + 1 \) . When solving this for \( y \) , we must now remember that our choice of \( x \geq - 3 \) becomes \( y \geq - 3 \) after replacing \( x \) with \( y \) . We now solve for \( y \) .\n\n\[ x = {\left( y + 3\right) }^{2} + 1 \Rightarrow x - 1 = {\left( y + 3\right) }^{2} \Rightarrow y + 3 = \pm \sqrt{x - 1} \]\n\n\[ \Rightarrow \;y = - 3 \pm \sqrt{x - 1} \]\n\nSince we have chosen the restriction of \( y \geq - 3 \), we use the expression with the positive sign, \( y = - 3 + \sqrt{x - 1} \), so that the inverse function is \( {f}^{-1}\left( x\right) = \) \( - 3 + \sqrt{x - 1} \) .\nb) For the function \( g \), the graph shows that we can restrict \( g \) to \( x > 2 \) to obtain a one-to-one function. The inverse for this choice is given as follows.\n\nReplacing \( x \) and \( y \) in \( y = \frac{1}{{\left( x - 2\right) }^{2}} \) gives \( x = \frac{1}{{\left( y - 2\right) }^{2}} \), which we solve this for \( y \) under the condition \( y > 2 \) .\n\n\[ x = \frac{1}{{\left( y - 2\right) }^{2}} \Rightarrow {\left( y - 2\right) }^{2} = \frac{1}{x} \Rightarrow y - 2 = \pm \frac{1}{\sqrt{x}} \]\n\n\[ \Rightarrow y = 2 \pm \frac{1}{\sqrt{x}} \Rightarrow {g}^{-1}\left( x\right) = 2 + \frac{1}{\sqrt{x}} \]\nc) Finally, \( h\left( x\right) = {x}^{3} - 3{x}^{2} \) can be graphed with the calculator as follows: \n\nThe picture on the right shows that the approximation of the local minimum is (approximately) at \( x = 2 \) . Therefore, if we restrict \( h \) to all \( x \geq 2 \), we obtain a one-to-one function. We replace \( x \) and \( y \) in \( y = {x}^{3} - 3{x}^{2} \), so that the inverse function is obtained by solving the equation \( x = {y}^{3} - 3{y}^{2} \) for \( y \) . However, this equation is quite complicated and solving it is beyond our capabilities at this time. Therefore we simply say that \( {h}^{-1}\left( x\right) \) is that \( y \geq 2 \) for which \( {y}^{3} - 3{y}^{2} = x \), and leave the example with this.
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Yes
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Are the following functions inverse to each other?
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Solution. We calculate the compositions \( f\left( {g\left( x\right) }\right) \) and \( g\left( {f\left( x\right) }\right) \) .\n\n\[ \text{a)}\;f\left( {g\left( x\right) }\right) = f\left( \frac{x - 7}{5}\right) = 5 \cdot \frac{x - 7}{5} + 7 = \left( {x - 7}\right) + 7 = x\text{,}\]\n\n\[ g\left( {f\left( x\right) }\right) = g\left( {{5x} + 7}\right) = \frac{\left( {{5x} + 7}\right) - 7}{5} = \frac{5x}{5} = x,\]\n\n\[ \text{b)}f\left( {g\left( x\right) }\right) = f\left( {\frac{3}{x} + 6}\right) = \frac{3}{\left( {\frac{3}{x} + 6}\right) - 6} = \frac{3}{\frac{3}{x}} = 3 \cdot \frac{x}{3} = x\text{,}\]\n\n\[ g\left( {f\left( x\right) }\right) = g\left( \frac{3}{x - 6}\right) = \frac{3}{\frac{3}{x - 6}} + 6 = 3 \cdot \frac{x - 6}{3} + 6 = \left( {x - 6}\right) + 6 = x.\]
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Yes
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Example 7.13. Find the graph of the inverse function of the function given below.\n\nb) \( f\left( x\right) = {\left( x + 1\right) }^{3} \)
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Solution. Carefully reflecting the graphs given in part (a) and (b) gives the following solution. The function \( f\left( x\right) = {\left( x + 1\right) }^{3} \) in part (b) can be graphed with a calculator first.
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No
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Divide the following fractions via long division:\n\n\[ \n\\text{a)}\\frac{3571}{11}\\;\\text{b)}\\frac{{x}^{3} + 5{x}^{2} + {4x} + 2}{x + 3} \n\]
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Solution. a) Recall the procedure for long division of natural numbers: \n\nThe steps above are performed as follows. First, we find the largest multiple of 11 less or equal to 35 . The answer 3 is written as the first digit on the top line. Multiply 3 times 11, and subtract the answer 33 from the first two digits 35 of the dividend. The remaining digits 71 are copied below to give 271. Now we repeat the procedure, until we arrive at the remainder 7. In short, what we have shown is that:\n\n\[ \n{3571} = {324} \\cdot {11} + 7\\;\\text{ or alternatively,}\\;\\frac{3571}{11} = {324} + \\frac{7}{11}. \n\]
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No
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Divide the following fractions via long division.
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Solution. For part (a), we calculate:\n\n\n\nTherefore, \( {x}^{2} + {4x} + 5 = \\left( {x + 8}\\right) \\cdot \\left( {x - 4}\\right) + {37} \) .
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No
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Find the remainder of dividing \( f\left( x\right) = {x}^{2} + {3x} + 2 \) by\n\n\[ \text{a)}x - 3\text{, b)}x + 4\text{, c)}x + 1\text{, d)}x - \frac{1}{2}\text{.} \]
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Solution. By Observation 8.10 we know that the remainder \( r \) of the division by \( x - c \) is \( f\left( c\right) \) . Thus, the remainder for part (a), when dividing by \( x - 3 \) is\n\n\[ r = f\left( 3\right) = {3}^{2} + 3 \cdot 3 + 2 = 9 + 9 + 2 = {20}. \]\n\nFor (b), note that \( g\left( x\right) = x + 4 = x - \left( {-4}\right) \), so that we take \( c = - 4 \) for our input giving that \( r = f\left( {-4}\right) = {\left( -4\right) }^{2} + 3 \cdot \left( {-4}\right) + 2 = {16} - {12} + 2 = 6 \) . Similarly, the other remainders are:\n\n\[ \text{c)}r = f\left( {-1}\right) = {\left( -1\right) }^{2} + 3 \cdot \left( {-1}\right) + 2 = 1 - 3 + 2 = 0\text{,}\]\n\n\[ \text{d)}r = f\left( \frac{1}{2}\right) = {\left( \frac{1}{2}\right) }^{2} + \frac{1}{2} \cdot 3 + 2 = \frac{1}{4} + \frac{3}{2} + 2 = \frac{1 + 6 + 8}{4} = \frac{15}{4}\text{.} \]
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Yes
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Determine whether \( g\left( x\right) \) is a factor of \( f\left( x\right) \). \[ \text{a)}f\left( x\right) = {x}^{3} + 2{x}^{2} + {5x} + 1,\;g\left( x\right) = x - 2\text{,} \]
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Solution. a) We need to determine whether 2 is a root of \( f\left( x\right) = {x}^{3} + 2{x}^{2} + {5x} + 1 \), that is, whether \( f\left( 2\right) \) is zero. \[ f\left( 2\right) = {2}^{3} + 2 \cdot {2}^{2} + 5 \cdot 2 + 1 = 8 + 8 + {10} + 1 = {27}. \] Since \( f\left( 2\right) = {27} \neq 0 \), we see that \( g\left( x\right) = x - 2 \) is not a factor of \( f\left( x\right) \).
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Yes
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a) Show that -2 is a root of \( f\left( x\right) = {x}^{5} - 3 \cdot {x}^{3} + 5{x}^{2} - {12} \), and use this to factor \( f \) .
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a) First, we calculate that -2 is a root.\n\n\[ f\left( {-2}\right) = {\left( -2\right) }^{5} - 3 \cdot {\left( -2\right) }^{3} + 5 \cdot {\left( -2\right) }^{2} + {12} = - {32} + {24} + {20} - {12} = 0. \]\n\nSo we can divide \( f\left( x\right) \) by \( g\left( x\right) = x - \left( {-2}\right) = x + 2 \) :\n\n\[ \begin{array}{r} - \left( {-2{x}^{4}\; - 4{x}^{3}}\right) \\ {x}^{3} \end{array} \]\n\nSo we factored \( f\left( x\right) \) as \( f\left( x\right) = \left( {{x}^{4} - 2{x}^{3} + {x}^{2} + {3x} - 6}\right) \cdot \left( {x + 2}\right) \) .
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Yes
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Our first example is the long division of \( \frac{5{x}^{3} + 7{x}^{2} + x + 4}{x + 2} \) .
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Here, the first term \( 5{x}^{2} \) of the quotient is just copied from the first term of the dividend. We record this together with the coefficients of the dividend \( 5{x}^{3} + 7{x}^{2} + x + 4 \) and of the divisor \( x + 2 = x - \left( {-2}\right) \) as follows:\n\n\[ - 2\left| {\;\begin{array}{llll} 5 & 7 & 1 & 4 \\ & & & \\ 5 & & & \end{array}}\right. \;\begin{array}{l} \text{ (dividend }\left( {5{x}^{3} + 7{x}^{2} + x + 4}\right) ) \\ \text{ (divisor }\left( {x - \left( {-2}\right) }\right) ) \\ \text{ (quotient) } \end{array} \]\n\nThe first actual calculation is performed when multiplying the \( 5{x}^{2} \) term with 2, and subtracting it from \( 7{x}^{2} \) . We record this as follows.\n\n\[ - 2\left| \begin{matrix} 5 & 7 & 1 & 4 \\ & - {10} & & \\ 5 & - 3 & & \left( {-3\text{ is the product of } - 2 \cdot 5}\right) \\ 5 & - 3 & & \left( {-3\text{ is the sum }7 + \left( {-{10}}\right) }\right) \end{matrix}\right| \]\n\nSimilarly, we obtain the next step by multiplying the \( {2x} \) by \( \left( {-3}\right) \) and subtracting it from \( {1x} \) . Therefore, we get\n\n\[ - 2\left| {\;\begin{matrix} 5 & 7 & 1 & 4 \\ & - {10} & 6 & \\ 5 & - 3 & 7 & \\ & & & \end{matrix}\;\left( {6\text{ is the product of } - 2 \cdot \left( {-3}\right) }\right) }\right. \]\n\nThe last step multiplies 7 times 2 and subtracts this from 4. In short, we write:\n\nThe answer can be determined from these coefficients. The quotient is \( 5{x}^{2} - \) \( {3x} + 7 \), and the remainder is -10 .
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Yes
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Example 8.15. Find the following quotients via synthetic division.\n\n\\[ \n\\text{a)}\\frac{4{x}^{3} - 7{x}^{2} + {4x} - 8}{x - 4}\\text{, b)}\\frac{{x}^{4} - {x}^{2} + 5}{x + 3}\\text{.} \n\\]
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Solution. a) We need to perform the synthetic division. \n\nTherefore we have\n\n\\[ \n\\frac{4{x}^{3} - 7{x}^{2} + {4x} - 8}{x - 4} = 4{x}^{2} + {9x} + {40} + \\frac{152}{x - 4}.\n\\]
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Yes
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Example 9.8. Which of the following graphs could be the graphs of a polynomial? If the graph could indeed be a graph of a polynomial then determine a possible degree of the polynomial.
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## Solution.\na) Yes, this could be a polynomial. The degree could be, for example, 4.\nb) No, since the graph has a pole.\nc) Yes, this could be a polynomial. A possible degree would be degree 3 .\nd) No, since the graph has a corner.\ne) No, since \( f\left( x\right) \) does not approach \( \infty \) or \( - \infty \) as \( x \) approaches \( \infty \) . (In fact \( f\left( x\right) \) approaches 0 as \( x \) approaches \( \pm \infty \) and we say that the function (or graph) has a horizontal asymptote \( y = 0 \) .)
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Yes
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Identify the graphs of the polynomials in (a), (b) and (c) with the functions (i), (ii), and (iii).
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Solution. The only odd degree function is (i) and it must therefore correspond to graph (c). For (ii), since the leading coefficient is negative, we know that the function opens downwards, so that it corresponds to graph (a). Finally (iii) opens upwards, since the leading coefficient is positive, so that its graph is \( \left( \mathrm{b}\right) \) .
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Yes
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Example 9.10. Graph of the given function with the TI-84. Include all extrema and intercepts of the graph in your viewing window.\na) \( f\left( x\right) = - {x}^{3} + {26}{x}^{2} - {129x} + {175} \)
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Solution. a) The graph in the standard window looks as follows:\n\n\n\nHowever, since the function is of degree 3 , this cannot be the full graph, as \( f\left( x\right) \) has to approach \( - \infty \) when \( x \) approaches \( \infty \) . Zooming out, and rescaling appropriately for the following setting the following graph.\n\n
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Yes
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Example 9.11. Find the roots of the polynomial from its graph.
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Solution. a) We start by graphing the polynomial \( f\left( x\right) = {x}^{3} - 7{x}^{2} + {14x} - 8 \) .\n\n\n\nThe graph suggests that the roots are at \( x = 1, x = 2 \), and \( x = 4 \) . This may easily be checked by looking at the function table.\n\n\n\nSince the polynomial is of degree 3, there cannot be any other roots.
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Yes
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The solutions of the equation \( a{x}^{2} + {bx} + c = 0 \) for some real numbers \( a, b \), and \( c \) are given by
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\[ {x}_{1} = \frac{-b + \sqrt{{b}^{2} - {4ac}}}{2a}\text{, and}{x}_{2} = \frac{-b - \sqrt{{b}^{2} - {4ac}}}{2a}\text{.} \] We may combine the two solutions \( {x}_{1} \) and \( {x}_{2} \) and simply write this as: \[ {x}_{1/2} = \frac{-b \pm \sqrt{{b}^{2} - {4ac}}}{2a} \]
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Yes
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Consider the equation \( {10}{x}^{3} - 6{x}^{2} + {5x} - 3 = 0 \) . Let \( x \) be a rational solution of this equation, that is \( x = \frac{p}{q} \) is a rational number such that\n\n\[ {10} \cdot {\left( \frac{p}{q}\right) }^{3} - 6 \cdot {\left( \frac{p}{q}\right) }^{2} + 5 \cdot \frac{p}{q} - 3 = 0. \]
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We assume that \( x = \frac{p}{q} \) is completely reduced, that is, \( p \) and \( q \) have no common factors that can be used to cancel the numerator and denominator of the fraction \( \frac{p}{q} \) . Now, simplifying the above equation, and combining terms, we obtain:\n\n\[ {10} \cdot \frac{{p}^{3}}{{q}^{3}} - 6 \cdot \frac{{p}^{2}}{{q}^{2}} + 5 \cdot \frac{p}{q} - 3 = 0 \]\n\n\[ \text{(multiply by}{q}^{3}\text{)} \Rightarrow {10}{p}^{3} - 6{p}^{2}q + {5p}{q}^{2} - 3{q}^{3} = 0 \]\n\n\[ \text{(add}3{q}^{3}\text{)} \Rightarrow {10}{p}^{3} - 6{p}^{2}q + {5p}{q}^{2} = 3{q}^{3} \]\n\n\[ \text{(factor}p\text{on the left)} \Rightarrow p \cdot \left( {{10}{p}^{2} - {6pq} + 5{q}^{2}}\right) = 3{q}^{3}\text{.} \]\n\nTherefore, \( p \) is a factor of \( 3{q}^{3} \) (with the other factor being \( \left( {{10}{p}^{2} - {6pq} + 5{q}^{2}}\right) \) ). Since \( p \) and \( q \) have no common factors, \( p \) must be a factor of 3 . That is, \( p \) is one of the following integers: \( p = + 1, + 3, - 1, - 3 \) .\n\nSimilarly starting from \( {10}{p}^{3} - 6{p}^{2}q + {5p}{q}^{2} - 3{q}^{3} = 0 \), we can write\n\n\[ \left( {\operatorname{add} + 6{p}^{2}q - {5p}{q}^{2} + 3{q}^{3}}\right) \Rightarrow {10}{p}^{3} = 6{p}^{2}q - {5p}{q}^{2} + 3{q}^{3} \]\n\n\[ \text{(factor}q\text{on the right)} \Rightarrow {10}{p}^{3} = \left( {6{p}^{2} - {5pq} + 3{q}^{2}}\right) \cdot q\text{.} \]\n\nNow, \( q \) must be a factor of \( {10}{p}^{3} \) . Since \( q \) and \( p \) have no common factors, \( q \) must be a factor of 10 . In other words, \( q \) is one of the following numbers: \( q = \) \( \pm 1, \pm 2, \pm 5, \pm {10} \) . Putting this together with the possibilities for \( p = \pm 1, \pm 3 \) , we see that all possible rational roots are the following:\n\n\[ \pm \frac{1}{1},\; \pm \frac{1}{2},\; \pm \frac{1}{5},\; \pm \frac{1}{10},\; \pm \frac{3}{1},\; \pm \frac{3}{2},\; \pm \frac{3}{5},\; \pm \frac{3}{10}. \]
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Yes
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a) Find all rational roots of \( f\left( x\right) = 7{x}^{3} + {x}^{2} + {7x} + 1 \) .
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a) If \( x = \frac{p}{a} \) is a rational root, then \( p \) is a factor of 1, that is \( p = \pm 1 \) , and \( q \) is a factor of \( 7 \), that is \( q = \pm 1, \pm 7 \) . The candidates for rational roots are therefore \( x = \pm \frac{1}{1}, \pm \frac{1}{7} \) . To see which of these candidates are indeed roots of \( f \) we plug these numbers into \( f \) via the calculator. We obtain the following:\n\n \n\nNote that we entered the \( x \) -value as a fraction \
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Yes
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Find roots of the given polynomial and use this information to factor the polynomial completely.
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a) In order to find a root, we use the graph to make a guess for one of the roots.\n\n\n\nThe graph suggests that the roots may be at \( x = - 2 \) and \( x = 3 \) . This is also supported by looking at the table for the function.\n\n\n\nWe check that these are roots by plugging the numbers into the function.\n\n\[ f\left( {-2}\right) = 2 \cdot {\left( -2\right) }^{3} - 8 \cdot {\left( -2\right) }^{2} - 6 \cdot \left( {-2}\right) + {36} = - {16} - {32} + {12} + {36} = 0, \]\n\n\[ f\left( 3\right) = 2 \cdot {3}^{3} - 8 \cdot {3}^{2} - 6 \cdot 3 + {36} = {54} - {72} - {18} + {36} = 0. \]\n\nBy the factor theorem we can divide \( f\left( x\right) = 2{x}^{3} - 8{x}^{2} - {6x} + {36} \), for example, by \( \left( {x - 3}\right) \) .\n\n\n\n\n\nTherefore, \( f\left( x\right) = \left( {2{x}^{2} - {2x} - {12}}\right) \left( {x - 3}\right) \) . To factor \( f \) completely, we still need to factor the first polynomial \( 2{x}^{2} - {2x} - {12} \) :\n\n\[ f\left( x\right) = \left( {2{x}^{2} - {2x} - {12}}\right) \left( {x - 3}\right) = 2 \cdot \left( {{x}^{2} - x - 6}\right) \cdot \left( {x - 3}\right) \]\n\n\[ = 2 \cdot \left( {x - 3}\right) \cdot \left( {x + 2}\right) \cdot \left( {x - 3}\right) = 2 \cdot \left( {x + 2}\right) \cdot {\left( x - 3\right) }^{2}. \]\n
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Yes
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Find the roots of the polynomial and sketch its graph including all roots.
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a) To find a root, we first graph the function \( f\left( x\right) = {x}^{3} + 2{x}^{2} - {14x} - 3 \) with the calculator. The graph and the table suggest that we have a root at \( x = 3 \). Therefore we divide \( f\left( x\right) \) by \( \left( {x - 3}\right) \). We obtain: This shows that \( f\left( x\right) = \left( {x - 3}\right) \left( {{x}^{2} + {5x} + 1}\right) \). To find the roots of \( f \), we also have to find the roots of the second factor \( {x}^{2} + {5x} + 1 \), i.e., the solutions to \( {x}^{2} + {5x} + 1 = 0 \). The quadratic formula gives: \[ x = \frac{-5 \pm \sqrt{{5}^{2} - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{-5 \pm \sqrt{21}}{2} \] The roots are therefore: \[ x = 3,\;x = \frac{-5 + \sqrt{21}}{2} \approx - {0.2},\;x = \frac{-5 - \sqrt{21}}{2} \approx - {4.8} \] We put these roots together with the graph in an appropriate window. The graph including the roots is displayed below.
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Yes
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Find a polynomial \( f \) with the following properties.\na) \( f \) has degree 3, the roots of \( f \) are precisely \( 4,5,6 \), and the leading coefficient of \( f \) is 7
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In general a polynomial \( f \) of degree 3 is of the form \( f\left( x\right) = \) \( m \cdot \left( {x - {c}_{1}}\right) \cdot \left( {x - {c}_{2}}\right) \cdot \left( {x - {c}_{3}}\right) \) . Identifying the roots and the leading coefficient, we obtain the polynomial\n\n\[ f\left( x\right) = 7 \cdot \left( {x - 4}\right) \cdot \left( {x - 5}\right) \cdot \left( {x - 6}\right) . \]
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Yes
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a) Our first graph is \( f\left( x\right) = \frac{1}{x - 3} \) .
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Here, the domain is all numbers where the denominator is not zero, that is \( D = \mathbb{R} - \{ 3\} \) . There is a vertical asymptote, \( x = 3 \) . Furthermore, the graph approaches 0 as \( x \) approaches \( \pm \infty \) . Therefore, \( f \) has a horizontal asymptote, \( y = 0 \) . Indeed, whenever the denominator has a higher degree than the numerator, the line \( y = 0 \) will be the horizontal asymptote.
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Yes
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Example 11.5. Find the domain, all horizontal asymptotes, vertical asymptotes, removable singularities, and \( x \) - and \( y \) -intercepts. Use this information together with the graph of the calculator to sketch the graph of \( f \) .
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Solution. a) We combine our knowledge of rational functions and its algebra with the particular graph of the function. The calculator gives the following graph.\n\n\n\nTo find the domain of \( f \) we only need to exclude from the real numbers those \( x \) that make the denominator zero. Since \( {x}^{2} - {3x} - 4 = 0 \) exactly when \( \left( {x + 1}\right) \left( {x - 4}\right) = 0 \), which gives \( x = - 1 \) or \( x = 4 \), we have the domain:\n\n\[ \text{domain}D = \mathbb{R} - \{ - 1,4\} \]\n\nThe numerator has a root exactly when \( - {x}^{2} = 0 \), that is \( x = 0 \) . Therefore, \( x = - 1 \) and \( x = 4 \) are vertical asymptotes, and since we cannot cancel terms in the fraction, there is no removable singularity. Furthermore, since \( f\left( x\right) = 0 \) exactly when the numerator is zero, the only \( x \) -intercept is \( \left( {0,0}\right) \) .\n\nTo find the horizontal asymptote, we consider \( f\left( x\right) \) for large values of \( x \) by ignoring the lower order terms in numerator and denominator,\n\n\[ \left| x\right| \text{ large } \Rightarrow f\left( x\right) \approx \frac{-{x}^{2}}{{x}^{2}} = - 1 \]\n\nWe see that the horizontal asymptote is \( y = - 1 \) . Finally, for the \( y \) -intercept, we calculate \( f\left( 0\right) \) :\n\n\[ f\left( 0\right) = \frac{-{0}^{2}}{{0}^{2} - 3 \cdot 0 - 4} = \frac{0}{-4} = 0. \]\n\nTherefore, the \( y \) -intercept is \( \left( {0,0}\right) \) . The function is then graphed as follows.\n\n
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Yes
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Graph the function \( p\left( x\right) = \frac{-3{x}^{2}{\left( x - 2\right) }^{3}\left( {x + 2}\right) }{\left( {x - 1}\right) {\left( x + 1\right) }^{2}{\left( x - 3\right) }^{3}} \) .
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Solution. We can see that \( p \) has zeros at \( x = 0,2 \), and -2 and vertical asymptotes \( x = 1 \) , \( x = - 1 \) and \( x = 3 \) . Also note that for large \( \left| x\right| \) , \( p\left( x\right) \approx - 3 \) . So there is a horizontal asymptote \( y = - 3 \) . We indicate each of these facts\non the graph: \n\nWe can in fact get a more precise statement by performing a long division and writing \( p\left( x\right) = \frac{n\left( x\right) }{d\left( x\right) } = \) \( - 3 + \frac{r\left( x\right) }{d\left( x\right) } \) . If you drop all but the leading order terms in the numerator and the denominator of the second term, we see that \( p\left( x\right) \approx - 3 - \frac{12}{x} \) whose graph for large \( \left| x\right| \) looks like\n\n\n\nThis sort of reasoning can make the graph a little more accurate but is not necessary for a sketch.\n\nWe also have the following table:\n\n<table><thead><tr><th>for \( a \)</th><th>near \( a \) , \( p\left( x\right) \approx \)</th><th>type</th><th>sign change at \( a \)</th></tr></thead><tr><td>\( - 2 \)</td><td>\( {C}_{1}\left( {x + 2}\right) \)</td><td>linear</td><td>changes</td></tr><tr><td>\( - 1 \)</td><td>\( {C}_{2}/{\left( x + 1\right) }^{2} \)</td><td>asymptote</td><td>does not change</td></tr><tr><td>0</td><td>\( {C}_{3}{x}^{2} \)</td><td>parabola</td><td>does not change</td></tr><tr><td>1</td><td>\( {C}_{4}/\left( {x - 1}\right) \)</td><td>asymptote</td><td>changes</td></tr><tr><td>2</td><td>\( {C}_{5}{\left( x - 2\right) }^{3} \)</td><td>cubic</td><td>changes</td></tr><tr><td>3</td><td>\( {C}_{6}/{\left( x - 3\right) }^{3} \)</td><td>asymptote</td><td>changes</td></tr></table>
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Yes
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Example 11.7. Sketch the graph of\n\n\[ r\left( x\right) = \frac{2{x}^{2}{\left( x - 1\right) }^{3}\left( {x + 2}\right) }{{\left( x + 1\right) }^{4}{\left( x - 2\right) }^{3}}. \]
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Solution. Here we see that there are \( x \) -intercepts at \( \left( {0,0}\right) ,\left( {0,1}\right) \), and \( \left( {0, - 2}\right) \). There are two vertical asymptotes: \( x = - 1 \) and \( x = 2 \). In addition, there is a horizontal asymptote at \( y = 0 \). (Why?) Putting this information on the graph gives\n\n\n\nIn this case, it is easy to get more information for large \( \left| x\right| \) that will be helpful in sketching the function. Indeed, when \( \left| x\right| \) is large, we can approximate \( r\left( x\right) \) by dropping all but the highest order term in the numerator and denominator which gives \( r\left( x\right) \approx \frac{2{x}^{6}}{{x}^{7}} = \frac{2}{x} \). So for large \( \left| x\right| \), the graph of \( r \) looks like\n\n
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Yes
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Solve for \( x \). a) \( -3x + 7 > 19 \), b) \( 2x + 5 \geq 4x - 11 \)
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Solution. The first three calculations are straightforward.\n\n\[ \text{a)}\; - {3x} + 7 > {19}\;\overset{\left( -7\right) }{ \Rightarrow }\; - {3x} > {12}\overset{\left( \div \left( -3\right) \right) }{ \Rightarrow }\;x < - 4 \]\n\n\[ \text{b)}\;{2x} + 5 \geq {4x} - {11}\;\overset{\left( -4x - 5\right) }{ \Rightarrow }\; - {2x} \geq - {16}\overset{\left( \div \left( -2\right) \right) }{ \Rightarrow }\;x \leq 8 \]
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Yes
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Example 12.2. Solve for \( x \). a) \( {x}^{2} - {3x} - 4 \geq 0 \)
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Solution. a) We can find the roots of the polynomial on the left by factoring.\n\n\[ {x}^{2} - {3x} - 4 = 0 \Rightarrow \;\left( {x - 4}\right) \left( {x + 1}\right) = 0 \Rightarrow \;x = 4\text{ or }x = - 1 \]\n\nTo see where \( f\left( x\right) = {x}^{2} - {3x} - 4 \) is \( \geq 0 \), we graph it with the calculator.\n\n\n\n\n\nWe see that \( f\left( x\right) \geq 0 \) when \( x \leq - 1 \) and when \( x \geq 4 \) (the parts of the graph above the \( x \) -axis). The solution set is therefore\n\n\[ \{ x \mid x \leq - 1,\text{ or }x \geq 4\} = \left( {-\infty , - 1\rbrack \cup \lbrack 4,\infty }\right) . \]
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Yes
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Find the domain of the given functions.\n\n\[ \text{a)}f\left( x\right) = \sqrt{{x}^{2} - 4}\;\text{b)}g\left( x\right) = \sqrt{{x}^{3} - 5{x}^{2} + {6x}} \]
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Solution. a) The domain of \( f\left( x\right) = \sqrt{{x}^{2} - 4} \) is given by all \( x \) for which the square root is non-negative. In other words the domain is given by numbers \( x \) with \( {x}^{2} - 4 \geq 0 \) . Graphing the function \( y = {x}^{2} - 4 = \left( {x + 2}\right) \left( {x - 2}\right) \), we see that this is precisely the case, when \( x \leq - 2 \) or \( x \geq 2 \) . \n\nTherefore, the domain is \( {D}_{f} = \left( {-\infty , - 2\rbrack \cup \lbrack 2,\infty }\right) \) .
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No
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a) \( \frac{{x}^{2} - {5x} + 6}{{x}^{2} - {5x}} \geq 0 \)
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Here is the graph of \( \frac{{x}^{2} - {5x} + 6}{{x}^{2} - {5x}} \) in the standard window.\n\n\n\nFactoring numerator and denominator, we can determine vertical asymptotes, holes, and \( x \) -intercepts.\n\n\[ \frac{{x}^{2} - {5x} + 6}{{x}^{2} - {5x}} = \frac{\left( {x - 2}\right) \left( {x - 3}\right) }{x\left( {x - 5}\right) } \]\n\nThe vertical asymptotes are at \( x = 0 \) and \( x = 5 \) , the \( x \) -intercepts are at \( x = 2 \) and \( x = 3 \) . Since for large \( x \), the fraction reduces to \( \frac{{x}^{2}}{{x}^{2}} = 1 \), we see that the horizontal asymptote is at \( y = 1 \) . Thus, \( \frac{{x}^{2} - {5x} + 6}{{x}^{2} - {5x}} \geq 0 \) for \( x < 0 \) and \( x > 5 \) . To see where the graph is \( \geq 0 \) between 0 and 5, we zoom into the graph:\n\n\n\nCombining all of the above information, we obtain the solution set:\n\n\[ \text{solution set} = \left( {-\infty ,0}\right) \cup \left\lbrack {2,3}\right\rbrack \cup \left( {5,\infty }\right) \]\n\nNotice that the \( x \) -coordinate of the \( x \) -intercepts are \( x = 2 \) and \( x = 3 \) are included in the solution set, whereas the values \( x = 0 \) and \( x = 5 \) associated with the vertical asymptotes are not included since the fraction is not defined for \( x = 0 \) and \( x = 5 \) .
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Yes
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Graph the functions \[ f\left( x\right) = {2}^{x},\;g\left( x\right) = {3}^{x},\;h\left( x\right) = {10}^{x},\;k\left( x\right) = {\left( \frac{1}{2}\right) }^{x},\;l\left( x\right) = {\left( \frac{1}{10}\right) }^{x}. \]
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First, we will graph the function \( f\left( x\right) = {2}^{x} \) by calculating the function values in a table and then plotting the points in the \( x - y \) -plane. We can calculate the values by hand, or simply use the table function of the calculator to find the function values.\n\n\[ f\left( 0\right) = {2}^{0} = 1 \]\n\[ f\left( 1\right) = {2}^{1} = 2 \]\n\[ f\left( 2\right) = {2}^{2} = 4 \]\n\[ f\left( 3\right) = {2}^{3} = 8 \]\n\[ f\left( {-1}\right) = {2}^{-1} = {0.5} \]\n\[ f\left( {-2}\right) = {2}^{-2} = {0.25} \]\n\nSimilarly, we can calculate the table for the other functions \( g, h, k \) and \( l \) by entering the functions in the spots at Y2, Y3, Y4, and Y5. The values in the table for these functions can be seen by moving the cursor to the right with the \( \left( \vartriangleright \right) \) key.\n\nWe can see the graphs by pressing the graph key.
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Yes
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Example 13.5. Graph the functions.\n\n\\[ \n\\text{a)}y = {e}^{x}\\text{, b)}y = {e}^{-x}\\text{, c)}y = {e}^{-{x}^{2}}\\text{, d)}y = \\frac{{e}^{x} + {e}^{-x}}{2} \n\\]
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Solution. Using the calculator, we obtain the desired graphs. The exponential function \\( y = {e}^{x} \\) may be entered via 2 nd ln\n\n\n\nNote that the minus sign is entered in the last expression (and also in the following two functions) via the \\( \\left( \\left( -\\right) \\right) \\) key.\n\n\n\nThe last function \\( y = \\frac{{e}^{x} + {e}^{-x}}{2} \\) is called the hyperbolic cosine, and is denoted by \\( \\cosh \\left( x\\right) = \\frac{{e}^{x} + {e}^{-x}}{2} \\) .
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Yes
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Example 13.6. Graph the functions.\n\n\[ \n\\text{a)}y = {2}^{x}\\text{, b)}y = 3 \\cdot {2}^{x}\\text{, c)}y = \\left( {-3}\\right) \\cdot {2}^{x}\n\]\n\n\[ \n\\text{d)}y = {0.2} \\cdot {2}^{x}\\text{, e)}y = \\left( {-{0.2}}\\right) \\cdot {2}^{x}\n\]
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Solution. We graph the functions in one viewing window.\n\n\n\n\n\nHere are the graphs of functions \( f\\left( x\\right) = c \\cdot {2}^{x} \) for various choices of \( c \) .\n\n\n\nNote that for \( f\\left( x\\right) = c \\cdot {2}^{x} \), the \( y \) -intercept is given at \( f\\left( 0\\right) = c \) .
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Yes
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Graph the functions. \[ \text{a)}y = {3}^{x} - 5\text{, b)}y = {e}^{x + 4}\text{, c)}y = \frac{1}{4} \cdot {e}^{x - 3} + 2 \]
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Solution. The graphs are displayed below. The first graph \( y = {3}^{x} - 5 \) is the graph of \( y = {3}^{x} \) shifted down by 5.
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No
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Rewrite the equation as a logarithmic equation.
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\[ \text{a)}{3}^{4} = {81}\text{, b)}{10}^{3} = {1000}\text{,} \] \[ \text{c)}{e}^{x} = {17}\text{, d)}{2}^{7 \cdot a} = {53}\text{.} \] Solution. We can immediately apply equation (13.2). For part (a), we have \( b = 3, y = 4 \), and \( x = {81} \) . Therefore we have: \[ {3}^{4} = {81}\; \Leftrightarrow \;{\log }_{3}\left( {81}\right) = 4 \] Similarly, we obtain the solutions for (b), (c), and (d). \[ \text{b)}{10}^{3} = {1000}\; \Leftrightarrow \;\log \left( {1000}\right) = 3 \] \[ \text{c)}{e}^{x} = {17}\; \Leftrightarrow \;\ln \left( {17}\right) = x \] \[ \text{d)}\;{2}^{7a} = {53}\; \Leftrightarrow \;{\log }_{2}\left( {53}\right) = {7a} \]
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Yes
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Evaluate the expression by rewriting it as an exponential expression.\na) \( {\log }_{2}\left( {16}\right) \)
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a) If we set \( y = {\log }_{2}\left( {16}\right) \), then this is equivalent to \( {2}^{y} = {16} \) . Since, clearly, \( {2}^{4} = {16} \), we see that \( y = 4 \) . Therefore, we have \( {\log }_{2}\left( {16}\right) = 4 \) .
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Yes
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Example 13.12. Calculate: a) \( {\log }_{3}\left( {13}\right) \)
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Solution. We calculate \( {\log }_{3}\left( {13}\right) \) by using the first formula in (13.4).\n\n\[ \n{\log }_{3}\left( {13}\right) = \frac{\log \left( {13}\right) }{\log \left( 3\right) } \approx {2.335} \n\] \n\nAlternatively, we can also calculate this with the second formula in (13.4).\n\n\[ \n{\log }_{3}\left( {13}\right) = \frac{\ln \left( {13}\right) }{\ln \left( 3\right) } \approx {2.335} \n\]
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Yes
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a) Graph the functions \( f\left( x\right) = \ln \left( x\right), g\left( x\right) = \log \left( x\right), h\left( x\right) = {\log }_{2}\left( x\right) \), and \( k\left( x\right) = {\log }_{0.5}\left( x\right) \) . What are the domains of \( f, g, h \), and \( k \) ? How do these functions differ?
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Solution. a) We know from the definition that the domain of \( f, g \), and \( h \) is all real positive numbers, \( {D}_{f} = {D}_{g} = {D}_{h} = {D}_{k} = \{ x \mid x > 0\} \) . The functions \( f \) and \( g \) can immediately be entered into the calculator. The standard window gives the following graphs.\n\n\n\nNote that we can rewrite \( g\left( x\right), h\left( x\right) \), and \( k\left( x\right) \) as a constant times \( f\left( x\right) \) :\n\n\[ g\left( x\right) = \log \left( x\right) = {\log }_{10}\left( x\right) = \frac{\ln \left( x\right) }{\ln \left( {10}\right) } = \frac{1}{\ln \left( {10}\right) } \cdot f\left( x\right) \]\n\n\[ h\left( x\right) = {\log }_{2}\left( x\right) = \frac{\ln \left( x\right) }{\ln \left( 2\right) } = \frac{1}{\ln \left( 2\right) } \cdot f\left( x\right) \]\n\n\[ k\left( x\right) = {\log }_{0.5}\left( x\right) = \frac{\ln \left( x\right) }{\ln \left( {0.5}\right) } = \frac{1}{\ln \left( {0.5}\right) } \cdot f\left( x\right) \]\n\nSince \( \frac{1}{\ln \left( {10}\right) } \approx {0.434} < 1 \), we see that the graph of \( g \) is that of \( f \) compressed towards the \( x \) -axis by a factor \( \frac{1}{\ln \left( {10}\right) } \) . Similarly, \( \frac{1}{\ln \left( 2\right) } \approx {1.443} > 1 \), so that the graph of \( h \) is that of \( f \) stretched away from the \( x \) -axis by a factor \( \frac{1}{\ln \left( 2\right) } \) . Finally, \( \frac{1}{\ln \left( {0.5}\right) } \approx - {1.443} \), or more precisely, \( \frac{1}{\ln \left( {0.5}\right) } = \frac{1}{\ln \left( {2}^{-1}\right) } = - \frac{1}{\ln \left( 2\right) } \), so that the graph of \( k \) is that of \( h \) reflected about the \( x \) -axis.\n\n\n\nNote that all these graphs have a common \( x \) -intercept at \( x = 1 \) :\n\n\[ f\left( 1\right) = g\left( 1\right) = h\left( 1\right) = k\left( 1\right) = 0 \]\n\nTo visualize the differences between the graphs, we graph them together in one coordinate system.\n\n
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Yes
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Proposition 14.1. The logarithm behaves well with respect to products, quotients, and exponentiation. Indeed, for all positive real numbers \( 0 < b \neq 1 \) , \( x > 0, y > 0 \), and real numbers \( n \), we have:\n\n\[{\log }_{b}\left( {x \cdot y}\right) = {\log }_{b}\left( x\right) + {\log }_{b}\left( y\right)\]\n\n\[{\log }_{b}\left( \frac{x}{y}\right) = {\log }_{b}\left( x\right) - {\log }_{b}\left( y\right)\]\n\n\[{\log }_{b}\left( {x}^{n}\right) = n \cdot {\log }_{b}\left( x\right)\]
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Proof. We start with the first formula \( {\log }_{b}\left( {x \cdot y}\right) = {\log }_{b}\left( x\right) + {\log }_{b}\left( y\right) \) . If we call \( u = {\log }_{b}\left( x\right) \) and \( v = {\log }_{b}\left( y\right) \), then the equivalent exponential formulas are \( {b}^{u} = x \) and \( {b}^{v} = y \) . With this, we have\n\n\[x \cdot y = {b}^{u} \cdot {b}^{v} = {b}^{u + v}.\n\nRewriting this in logarithmic form, we obtain\n\n\[{\log }_{b}\left( {x \cdot y}\right) = u + v = {\log }_{b}\left( x\right) + {\log }_{b}\left( y\right) .\n\nThis is what we needed to show.\n\nNext, we prove the formula \( {\log }_{b}\left( \frac{x}{y}\right) = {\log }_{b}\left( x\right) - {\log }_{b}\left( y\right) \) . We abbreviate \( u = {\log }_{b}\left( x\right) \) and \( v = {\log }_{b}\left( y\right) \) as before, and their exponential forms are \( {b}^{u} = x \) and \( {b}^{v} = y \) . Therefore, we have\n\n\[\frac{x}{y} = \frac{{b}^{u}}{{b}^{v}} = {b}^{u - v}.\n\nRewriting this again in logarithmic form, we obtain the desired result.\n\n\[{\log }_{b}\left( \frac{x}{y}\right) = u - v = {\log }_{b}\left( x\right) - {\log }_{b}\left( y\right)\n\nFor the third formula, \( {\log }_{b}\left( {x}^{n}\right) = n \cdot {\log }_{b}\left( x\right) \), we write \( u = {\log }_{b}\left( x\right) \), that is in exponential form \( {b}^{u} = x \).\n\nThen:\n\n\[{x}^{n} = {\left( {b}^{u}\right) }^{n} = {b}^{n \cdot u}\; \Rightarrow \;{\log }_{b}\left( {x}^{n}\right) = n \cdot u = n \cdot {\log }_{b}\left( x\right)\]
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Yes
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Combine the terms using the properties of logarithms so as to write as one logarithm.
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a) \( \frac{1}{2}\ln \left( x\right) + \ln \left( y\right) = \ln \left( {x}^{\frac{1}{2}}\right) + \ln \left( y\right) = \ln \left( {{x}^{\frac{1}{2}}y}\right) = \ln \left( {\sqrt{x} \cdot y}\right) \)
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Yes
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Example 14.3. Write the expressions in terms of elementary logarithms \( u = \) \( {\log }_{b}\left( x\right), v = {\log }_{b}\left( y\right) \), and, in part (c), also \( w = {\log }_{b}\left( z\right) \) . Assume that \( x, y, z > 0 \) .\n\n\[ \n\text{a)}\ln \left( {\sqrt{{x}^{5}} \cdot {y}^{2}}\right) ,\;\text{b)}\log \left( \sqrt{\sqrt{x} \cdot {y}^{3}}\right) \;\text{c)}{\log }_{2}\left( \sqrt[3]{\frac{{x}^{2}}{y\sqrt{z}}}\right) \text{.} \n\]
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Solution. In a first step, we rewrite the expression with fractional exponents, and then apply the rules from proposition 14.1\n\n\[ \n\text{a)}\ln \left( {\sqrt{{x}^{5}} \cdot {y}^{2}}\right) = \ln \left( {{x}^{\frac{5}{2}} \cdot {y}^{2}}\right) = \ln \left( {x}^{\frac{5}{2}}\right) + \ln \left( {y}^{2}\right) \n\]\n\n\[ \n= \frac{5}{2}\ln \left( x\right) + 2\ln \left( y\right) = \frac{5}{2}u + {2v} \n\]\n\n\[ \n\text{b)}\log \left( \sqrt{\sqrt{x} \cdot {y}^{3}}\right) = \log \left( {\left( {x}^{\frac{1}{2}}{y}^{3}\right) }^{\frac{1}{2}}\right) = \frac{1}{2}\log \left( {{x}^{\frac{1}{2}}{y}^{3}}\right) \n\]\n\n\[ \n= \frac{1}{2}\left( {\log \left( {x}^{\frac{1}{2}}\right) + \log \left( {y}^{3}\right) }\right) = \frac{1}{2}\left( {\frac{1}{2}\log \left( x\right) + 3\log \left( y\right) }\right) \n\]\n\n\[ \n= \frac{1}{4}\log \left( x\right) + \frac{3}{2}\log \left( y\right) = \frac{1}{4}u + \frac{3}{2}v \n\]\n\n\[ \n\text{c)}{\log }_{2}\left( \sqrt[3]{\frac{{x}^{2}}{y\sqrt{z}}}\right) = {\log }_{2}\left( {\left( \frac{{x}^{2}}{y \cdot {z}^{\frac{1}{2}}}\right) }^{\frac{1}{3}}\right) = \frac{1}{3}{\log }_{2}\left( \frac{{x}^{2}}{y \cdot {z}^{\frac{1}{2}}}\right) \n\]\n\n\[ \n= \frac{1}{3}\left( {{\log }_{2}\left( {x}^{2}\right) - {\log }_{2}\left( y\right) - {\log }_{2}\left( {z}^{\frac{1}{2}}\right) }\right) \n\]\n\n\[ \n= \frac{1}{3}\left( {2{\log }_{2}\left( x\right) - {\log }_{2}\left( y\right) - \frac{1}{2}{\log }_{2}\left( z\right) }\right) \n\]\n\n\[ \n= \frac{2}{3}{\log }_{2}\left( x\right) - \frac{1}{3}{\log }_{2}\left( y\right) - \frac{1}{6}{\log }_{2}\left( z\right) \n\]\n\n\[ \n= \frac{2}{3}u - \frac{1}{3}v - \frac{1}{6}w \n\]
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Yes
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Example 14.5. Solve for \( x \). a) \( {2}^{x + 7} = {32} \)
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\[ \text{a)}\;{2}^{x + 7} = {32}\; \Rightarrow \;{2}^{x + 7} = {2}^{5} \Rightarrow \;x + 7 = 5 \Rightarrow \;x = - 2 \]
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Yes
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Example 14.6. Solve for \( x \). a) \( {3}^{x + 5} = 8 \) b) \( {13}^{{2x} - 4} = 6 \) c) \( {5}^{x - 7} = {2}^{x} \) d) \( {5.1}^{x} = {2.7}^{{2x} + 6} \) e) \( {17}^{x - 2} = {3}^{x + 4} \) f) \( {7}^{{2x} + 3} = {11}^{{3x} - 6} \)
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Solution. We solve these equations by applying a logarithm (both log or ln will work for solving the equation), and then we use the identity \( \log \left( {a}^{x}\right) = \) \( x \cdot \log \left( a\right) \). a) \( {3}^{x + 5} = 8 \Rightarrow \ln {3}^{x + 5} = \ln 8 \Rightarrow \left( {x + 5}\right) \cdot \ln 3 = \ln 8 \Rightarrow x + 5 = \frac{\ln 8}{\ln 3} \Rightarrow x = \frac{\ln 8}{\ln 3} - 5 \approx - {3.11} \)
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No
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Let \( f\left( x\right) = c \cdot {b}^{x} \). Determine the constant \( c \) and base \( b \) under the given conditions.\na) \( f\left( 0\right) = 5,\;f\left( 1\right) = {20} \)
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Solution. a) Applying \( f\left( 0\right) = 5 \) to \( f\left( x\right) = c \cdot {b}^{x} \), we get\n\n\[ 5 = f\left( 0\right) = c \cdot {b}^{0} = c \cdot 1 = c \]\n\nIndeed, in general, we always have \( f\left( 0\right) = c \) for any exponential function. The base \( b \) is then determined by substituting the second equation \( f\left( 1\right) = {20} \).\n\n\[ {20} = f\left( 1\right) = c \cdot {b}^{1} = 5 \cdot b\;\overset{\left( \div 5\right) }{ \Rightarrow }\;b = 4 \]\n\nTherefore, \( f\left( x\right) = 5 \cdot {4}^{x} \). Note that in the last implication, we used that the base must be positive.
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Yes
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a) What is the mass of the bacteria sample after 4 hours?
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a) The formula for the mass \( y \) in grams after \( t \) hours is \( y\left( t\right) = 2 \cdot {1.02}^{t} \) . Therefore, after 4 hours, the mass in grams is\n\n\[ y\left( 4\right) = 2 \cdot {1.02}^{4} \approx {2.16}. \]\n\n
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Yes
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a) Assuming an exponential growth for the population size, find the formula for the population depending on the year \( t \) .
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a) The growth is assumed to be exponential, so that \( y\left( t\right) = c \cdot {b}^{t} \) describes the population size depending on the year \( t \), where we set \( t = 0 \) corresponding to the year 2000 . Then the example describes \( y\left( 0\right) = c \) as \( c = {12.7} \), which we assume in units of millions. To find the base \( b \), we substitute the data of \( t = {10} \) and \( y\left( t\right) = {14.3} \) into \( y\left( t\right) = c \cdot {b}^{t} \) .\n\n\[ \n{14.3} = {12.7} \cdot {b}^{10} \Rightarrow \frac{14.3}{12.7} = {b}^{10} \Rightarrow {\left( \frac{14.3}{12.7}\right) }^{\frac{1}{10}} = {\left( {b}^{10}\right) }^{\frac{1}{10}} = b \n\]\n\n\[ \n\Rightarrow \;b = {\left( \frac{14.3}{12.7}\right) }^{\frac{1}{10}} \approx {1.012} \n\]\n\nThe formula for the population size is \( y\left( t\right) \approx {12.7} \cdot {1.012}^{t} \) .
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Yes
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The number of PCs that are sold in the U.S. in the year 2011 is approximately 350 million with a rate of growth of \( {3.6}\% \) per year. Assuming the rate stays constant over the next years, how many PCs will be sold in the year 2015?
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Solution. Since the rate of growth is \( r = {3.6}\% = {0.036} \), we obtain a base of \( b = 1 + r = {1.036} \), giving the number of PCs to be modeled by \( c{\left( {1.036}\right) }^{t} \) . If we set \( t = 0 \) for the year 2011, we find that \( c = {350} \), so the number of sales is given by \( y\left( t\right) = {350} \cdot {1.036}^{t} \) . Since the year 2015 corresponds to \( t = 4 \), we can calculate the number of sales in the year 2015 as\n\n\[ y\left( 4\right) = {350} \cdot {1.036}^{4} \approx {403}. \]\n\nApproximately 403 million PCs will be sold in the year 2015.
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Yes
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The size of an ant colony is decreasing at a rate of \( 1\% \) per month. How long will it take until the colony has reached \( {80}\% \) of its original size?
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Solution. Since \( r = - 1\% = - {0.01} \), we obtain the base \( b = 1 + r = 1 - {0.01} = \) 0.99 . We have a colony size of \( y\left( t\right) = c \cdot {0.99}^{t} \) after \( t \) months, where \( c \) is the original size. We need to find \( t \) so that the size is at \( {80}\% \) of its original size \( c \), that is, \( y\left( t\right) = {80}\% \cdot c = {0.8} \cdot c \) .\n\n\[ \n{0.8} \cdot c = c \cdot {0.99}^{t}\;\overset{\left( \div c\right) }{ \Rightarrow }\;{0.8} = {0.99}^{t}\; \Rightarrow \log \left( {0.8}\right) = \log \left( {0.99}^{t}\right) \n\]\n\n\[ \n\Rightarrow \;\log \left( {0.8}\right) = t \cdot \log \left( {0.99}\right) \n\]\n\n\[ \n\Rightarrow \;t = \frac{\log \left( {0.8}\right) }{\log \left( {0.99}\right) } \approx {22.2} \n\]\n\nAfter approximately 22.2 months, the ant colony has decreased to \( {80}\% \) of its original size.
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Yes
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The population size of a country is increasing at a rate of \( 4\% \) per year. How long does it take until the country has doubled its population size?
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The rate of change is \( r = 4\% = {0.04} \), so that population size is an exponential function with base \( b = 1 + {0.04} = {1.04} \) . Therefore, \( f\left( x\right) = c \cdot {1.04}^{x} \) denotes the population size, with \( c \) being the initial population in the year corresponding to \( x = 0 \) . In order for the population size to double, \( f\left( x\right) \) has to reach twice its initial size, that is:\n\n\[ f\left( x\right) = {2c}\; \Rightarrow \;c \cdot {1.04}^{x} = {2c}\;\overset{\left( \div c\right) }{ \Rightarrow }\;{1.04}^{x} = 2 \]\n\n\[ \Rightarrow \;\log \left( {1.04}^{x}\right) = \log \left( 2\right) \; \Rightarrow \;x\log \left( {1.04}\right) = \log \left( 2\right) \]\n\n\[ \Rightarrow \;x = \frac{\log \left( 2\right) }{\log \left( {1.04}\right) } \approx {17.7} \]\n\nIt will take about 17.7 years until the population size has doubled.
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Yes
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Consider the function \( f\left( x\right) = {200} \cdot {\left( \frac{1}{2}\right) }^{\frac{x}{7}} \). We calculate the function values \( f\left( x\right) \), for \( x = 0,7,{14},{21} \), and 28.
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\[ f\left( 0\right) = {200} \cdot {\left( \frac{1}{2}\right) }^{\frac{0}{7}} = {200} \cdot 1 = {200} \] \[ f\left( 7\right) = {200} \cdot {\left( \frac{1}{2}\right) }^{\frac{7}{7}} = {200} \cdot \frac{1}{2} = {100} \] \[ f\left( {14}\right) = {200} \cdot {\left( \frac{1}{2}\right) }^{\frac{14}{7}} = {200} \cdot \frac{1}{4} = {50} \] \[ f\left( {21}\right) = {200} \cdot {\left( \frac{1}{2}\right) }^{\frac{21}{7}} = {200} \cdot \frac{1}{8} = {25} \] \[ f\left( {28}\right) = {200} \cdot {\left( \frac{1}{2}\right) }^{\frac{28}{7}} = {200} \cdot \frac{1}{16} = {12.5} \]
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Yes
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a) Chromium-51 has a half-life of 27.7 days1. How much of 3 grams of chromium-51 will remain after 90 days?
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Solution. a) We use the above formula \( y = c \cdot {\left( \frac{1}{2}\right) }^{\frac{t}{h}} \), where \( c = 3 \) grams is the initial amount of chromium-51, \( h = {27.7} \) days is the half-life of chromium-51, and \( t = {90} \) days is time that the isotope decayed. Substituting these numbers into the formula for \( y \), we obtain:\n\n\[ y = 3 \cdot {\left( \frac{1}{2}\right) }^{\frac{90}{27.7}} \approx {0.316} \]\n\nTherefore, after 90 days, 0.316 grams of the chromium-51 is remaining.
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Yes
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We invest an initial amount of \( P = \$ {500} \) for 1 year at a rate of \( r = 6\% \) . The initial amount \( P \) is also called the principal. After 1 year, we receive the principal \( P \) together with the interest \( r \) . \( P \) generated from the principal. The final amount \( A \) after 1 year is therefore
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\[ A = \$ {500} + 6\% \cdot \$ {500} = \$ {500} \cdot \left( {1 + {0.06}}\right) = \$ {530}. \]
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Yes
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Determine the final amount received on an investment under the given conditions.\na) \$700, compounded monthly, at 4%, for 3 years
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Solution. We can immediately use the formula by substituting the given values. For part (a), we have \( P = {700}, n = {12}, r = 4\% = {0.04} \), and \( t = 3 \) . Therefore, we calculate\n\n\[ A = {700} \cdot {\left( 1 + \frac{0.04}{12}\right) }^{{12} \cdot 3} = {700} \cdot {\left( 1 + \frac{0.04}{12}\right) }^{36} \approx {789.09} \]\n\n
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Yes
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a) Find the amount \( P \) that needs to be invested at \( {4.275}\% \) compounded annually for 5 years to give a final amount of \( \$ {3000} \) . (This amount \( P \) is also called the present value of the future amount of \( \$ {3000} \) in 5 years.)
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a) We have the following data: \( r = {4.275}\% = {0.04275}, n = 1 \) , \( t = 5 \), and \( A = {3000} \) . We want to find the present value \( P \) . Substituting the given numbers into the appropriate formula, we can solve this for \( P \) .\n\n\[ \n{3000} = P \cdot {\left( 1 + \frac{0.04275}{1}\right) }^{1.5} \Rightarrow \;{3000} = P \cdot {\left( {1.04275}\right) }^{5} \n\]\n\n\[ \n\text{(divide} \Rightarrow {A}^{2} = \frac{{3000}^{2}}{{1201275}^{5}} \approx {2433.44} \n\]\n\nTherefore, if we invest \( \$ {2433.44} \) today under the given conditions, then this will be worth \( \$ {3000} \) in 5 years.
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Yes
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Find \( \sin \left( x\right) ,\cos \left( x\right) \), and \( \tan \left( x\right) \) for the angles\n\n\[ x = {30}^{ \circ },\;x = {45}^{ \circ },\;x = {60}^{ \circ },\;x = {90}^{ \circ },\;x = {0}^{ \circ }. \]
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Solution. We draw the angle \( x = {30}^{ \circ } \) in the plane, and use the special triangle to find a point on the terminal side of the angle. From this, we then read off the trigonometric functions. Here are the function values for \( {30}^{ \circ },{45}^{ \circ } \) , and \( {60}^{ \circ } \) .\n\n\[ \sin \left( {30}^{ \circ }\right) = \frac{b}{r} = \frac{1}{2} \]\n\n\n\n\[ \cos \left( {30}^{ \circ }\right) = \frac{a}{r} = \frac{\sqrt{3}}{2} \]\n\n\[ \tan \left( {30}^{ \circ }\right) = \frac{b}{a} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3} \]\n\n\[ \sin \left( {45}^{ \circ }\right) = \frac{b}{r} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \]\n\n\[ \cos \left( {45}^{ \circ }\right) = \frac{a}{r} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \]\n\n\[ \tan \left( {45}^{ \circ }\right) = \frac{b}{a} = \frac{1}{1} = 1 \]\n\n\n\n\[ \sin \left( {60}^{ \circ }\right) = \frac{b}{r} = \frac{\sqrt{3}}{2} \]\n\n\[ \cos \left( {60}^{ \circ }\right) = \frac{a}{r} = \frac{1}{2} \]\n\n\[ \tan \left( {60}^{ \circ }\right) = \frac{b}{a} = \frac{\sqrt{3}}{1} = \sqrt{3} \]\n\nHere are the trigonometric functions for \( {90}^{ \circ } \) and \( {0}^{ \circ } \).\n\n\[ \sin \left( {90}^{ \circ }\right) = \frac{b}{r} = \frac{1}{1} = 1 \]\n\n\n\n\[ \cos \left( {90}^{ \circ }\right) = \frac{a}{r} = \frac{0}{1} = 0 \]\n\n\( \tan \left( {90}^{ \circ }\right) = \frac{b}{a} = \frac{1}{0} \) is undefined\n\n\[ \sin \left( {0}^{ \circ }\right) = \frac{b}{r} = \frac{0}{1} = 0 \]\n\n\[ \cos \left( {0}^{ \circ }\right) = \frac{a}{r} = \frac{1}{1} = 1 \]\n\n\[ \tan \left( {0}^{ \circ }\right) = \frac{b}{a} = \frac{0}{1} = 0 \]
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Yes
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Find \( \sin \left( x\right) ,\cos \left( x\right) \), and \( \tan \left( x\right) \) for the following angles.\n\n\[ \text{a)}x = {240}^{ \circ }\text{，b)}x = {495}^{ \circ }\text{，c)}x = \frac{11\pi }{6}\text{，d)}x = \frac{-{9\pi }}{4} \]
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Solution. a) We graph the angle \( x = {240}^{ \circ } \), and identify a special \( {30}^{ \circ } - {60}^{ \circ } - \) \( {90}^{ \circ } \) triangle using the terminal side of \( x \) .\n\n\n\nWe identify a point \( P \) on the terminal side of \( x \) . This point \( P \) has coordinates \( P\left( {-1, - \sqrt{3}}\right) \), which can be seen by considering one of two \( {30}^{ \circ } - {60}^{ \circ } - {90}^{ \circ } \) triangles in the plane (shaded above). The length of the line segment from \( P \) to the origin \( \left( {0,0}\right) \) is 2 . Thus, \( a = - 1, b = - \sqrt{3}, r = 2 \) . We get\n\n\[ \sin \left( {240}^{ \circ }\right) = \frac{-\sqrt{3}}{2},\;\cos \left( {240}^{ \circ }\right) = \frac{-1}{2},\;\tan \left( {240}^{ \circ }\right) = \frac{-\sqrt{3}}{-1} = \sqrt{3} \]
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No
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We graph the functions \( y = \sin \left( x\right), y = \cos \left( x\right) \), and \( y = \tan \left( x\right) \) .
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One way to proceed is to calculate and collect various function values in a table and then graph them. However, this is quite elaborate, as the following table shows.\n\n<table><tr><td>\( x \)</td><td>0</td><td>\( \frac{\pi }{6} \)</td><td>\( \frac{\pi }{4} \)</td><td>\( \frac{\pi }{3} \)</td><td>\( \frac{\pi }{2} \)</td><td>\( \frac{2\pi }{3} \)</td><td>\( \frac{3\pi }{4} \)</td><td>\( \frac{5\pi }{6} \)</td><td>\( \pi \)</td><td>\( \cdots \)</td></tr><tr><td>\( \sin \left( x\right) \)</td><td>0</td><td>\( \frac{1}{2} \)</td><td>\( \frac{\sqrt{2}}{2} \)</td><td>\( \frac{\sqrt{3}}{2} \)</td><td>1</td><td>\( \frac{\sqrt{3}}{2} \)</td><td>\( \frac{\sqrt{2}}{2} \)</td><td>\( \frac{1}{2} \)</td><td>0</td><td>...</td></tr><tr><td>\( \cos \left( x\right) \)</td><td>1</td><td>\( \frac{\sqrt{3}}{2} \)</td><td>\( \frac{\sqrt{2}}{2} \)</td><td>\( \frac{1}{2} \)</td><td>0</td><td>\( \frac{-1}{2} \)</td><td>\( \frac{-\sqrt{2}}{2} \)</td><td>\( \frac{-\sqrt{3}}{2} \)</td><td>\( - 1 \)</td><td>\( \cdots \)</td></tr><tr><td>\( \tan \left( x\right) \)</td><td>0</td><td>\( \frac{\sqrt{3}}{3} \)</td><td>1</td><td>\( \sqrt{3} \)</td><td>undef.</td><td>\( - \sqrt{3} \)</td><td>\( - 1 \)</td><td>\( \frac{-\sqrt{3}}{3} \)</td><td>0</td><td>...</td></tr></table>\n\nAn easier approach is to use the TI-84 to graph each function. Before entering the function, we need to make sure that the calculator is in radian mode. For this, press the mode key, and in case the third item is set on degree, switch to radian, and press enter ). \n\nWe may now enter the function \( y = \sin \left( x\right) \) and study its graph. \n\nFrom this, we can make some immediate observations (which can also be seen using definition 17.2). First, the graph is bounded between -1 and +1 , because by definition 17.2 we have \( \sin \left( x\right) = \frac{b}{r} \) with \( - r \leq b \leq r \), so that\n\n\[ - 1 \leq \sin \left( x\right) \leq 1,\;\text{ for all }x. \]\n\nWe therefore change the window of the graph to display \( y \) ’s between -2 and 2, to get a closer view.\n\n\n\nSecond, we see that \( y = \sin \left( x\right) \) is a periodic function with period \( {2\pi } \), since the function doesn’t change its value when adding \( {360}^{ \circ } = {2\pi } \) to its argument (and this is the smallest non-zero number with that property):\n\n\[ \sin \left( {x + {2\pi }}\right) = \sin \left( x\right) \]
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Yes
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Graph the functions:\n\n\[ f\left( x\right) = \sin \left( x\right) + 3, g\left( x\right) = 4 \cdot \sin \left( x\right), h\left( x\right) = \sin \left( {x + 2}\right), i\left( x\right) = \sin \left( {3x}\right) ,\nj\left( x\right) = 2 \cdot \cos \left( x\right) + 3, k\left( x\right) = \cos \left( {{2x} - \pi }\right), l\left( x\right) = \tan \left( {x + 2}\right) + 3 \]
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Solution. The functions \( f, g, h \), and \( i \) have graphs that are variations of the basic \( y = \sin \left( x\right) \) graph. The graph of \( f\left( x\right) = \sin \left( x\right) + 3 \), shifts the graph of \( y = \sin \left( x\right) \) up by 3, whereas the graph of \( g\left( x\right) = 4 \cdot \sin \left( x\right) \) stretches \( y = \sin \left( x\right) \) away from the \( x \) -axis.\n\n\n\nThe graph of \( h\left( x\right) = \sin \left( {x + 2}\right) \) shifts the graph of \( y = \sin \left( x\right) \) to the left by 2, and \( i\left( x\right) = \sin \left( {3x}\right) \) compresses \( y = \sin \left( x\right) \) towards the \( y \) -axis.\n\n
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No
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Example 17.10. Find the amplitude, period, and phase-shift, and sketch the graph over one full period.
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Solution. a) The amplitude is \( \\left| 3\\right| = 3 \), and since \( f\\left( x\\right) = 3 \\cdot \\sin \\left( {2 \\cdot x + 0}\\right) \), it is \( b = 2 \) and \( c = 0 \), so that the period is \( \\left| \\frac{2\\pi }{2}\\right| = \\pi \) and the phase-shift is \( \\frac{-0}{2} = 0 \) . This is also supported by graphing the function with the calculator; in particular the graph repeats after one full period, which is at \( \\pi \\approx {3.1} \) .
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Yes
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Proposition 18.1. For any angles \( \alpha \) and \( \beta \), we have\n\n\[ \sin \left( {\alpha + \beta }\right) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \]
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Proof. We start with the proof of the formulas for \( \sin \left( {\alpha + \beta }\right) \) and \( \cos \left( {\alpha + \beta }\right) \) when \( \alpha \) and \( \beta \) are angles between 0 and \( \frac{\pi }{2} = {90}^{ \circ } \) . We prove the addition formulas (for \( \alpha ,\beta \in \left( {0,\frac{\pi }{2}}\right) \) ) in a quite elementary way, and then show that the addition formulas also hold for arbitrary angles \( \alpha \) and \( \beta \) .\n\nTo find \( \sin \left( {\alpha + \beta }\right) \), consider the following setup.\n\n\n\nNote, that there are vertically opposite angles, labelled by \( \gamma \), which are therefore equal. These angles are angles in two right triangles, with the third angle being \( \alpha \) . We therefore see that the angle \( \alpha \) appears again as the angle among the sides \( b \) and \( f \) . With this, we can now calculate \( \sin \left( {\alpha + \beta }\right) \).\n\n\[ \sin \left( {\alpha + \beta }\right) = \frac{\text{ opposite }}{\text{ hypotenuse }} = \frac{e + f}{d} = \frac{e}{d} + \frac{f}{d} = \frac{a}{d} + \frac{f}{d} = \frac{a}{c} \cdot \frac{c}{d} + \frac{f}{b} \cdot \frac{b}{d} \]\n\n\[ = \sin \left( \alpha \right) \cos \left( \beta \right) + \cos \left( \alpha \right) \sin \left( \beta \right) \]\n\nThe above figure displays the situation when \( \alpha + \beta \leq \frac{\pi }{2} \) . There is a similar figure for \( \frac{\pi }{2} < \alpha + \beta < \pi \) . (We recommend as an exercise to draw the corresponding figure for the case of \( \frac{\pi }{2} < \alpha + \beta < \pi \) .)
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No
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Example 18.2. Find the exact values of the trigonometric functions.\n\n\\[ \n\\text{a)}\\cos \\left( \\frac{\\pi }{12}\\right) \\;\\text{b)}\\tan \\left( \\frac{5\\pi }{12}\\right) \\;\\text{c)}\\cos \\left( \\frac{11\\pi }{12}\\right) \n\\]
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Solution. a) The key is to realize the angle \\( \\frac{\\pi }{12} \\) as a sum or difference of angles with known trigonometric function values. Note, that \\( \\frac{\\pi }{3} - \\frac{\\pi }{4} = \\frac{{4\\pi } - {3\\pi }}{12} = \\frac{\\pi }{12} \\), so that\n\n\\[ \n\\cos \\left( \\frac{\\pi }{12}\\right) = \\cos \\left( {\\frac{\\pi }{3} - \\frac{\\pi }{4}}\\right) = \\cos \\frac{\\pi }{3}\\cos \\frac{\\pi }{4} + \\sin \\frac{\\pi }{3}\\sin \\frac{\\pi }{4} \n\\]\n\n\\[ \n= \\frac{1}{2} \\cdot \\frac{\\sqrt{2}}{2} + \\frac{\\sqrt{3}}{2} \\cdot \\frac{\\sqrt{2}}{2} = \\frac{\\sqrt{2}}{4} + \\frac{\\sqrt{6}}{4} = \\frac{\\sqrt{2} + \\sqrt{6}}{4} \n\\]\n\nWe remark that the last expression is in the simplest form and cannot be simplified any further.
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Yes
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We rewrite \( \cos \left( {x + \frac{\pi }{2}}\right) \) by using the subtraction formula.
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\n\[
\cos \left( {x + \frac{\pi }{2}}\right) = \cos x \cdot \cos \frac{\pi }{2} - \sin x \cdot \sin \frac{\pi }{2} = \cos x \cdot 0 + \sin x \cdot 1 = \sin \left( x\right)
\]
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Yes
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Find the trigonometric functions using the half-angle formulas.
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\[ \text{a)}\sin \left( \frac{\pi }{8}\right) \;\text{b)}\cos \left( \frac{9\pi }{8}\right) \;\text{c)}\tan \left( \frac{\pi }{24}\right) \] Solution. a) Since \( \frac{\pi }{8} = \frac{\frac{\pi }{4}}{2} \), we use the half-angle formula with \( \alpha = \frac{\pi }{4} \). \[ \sin \left( \frac{\pi }{8}\right) = \sin \left( \frac{\frac{\pi }{4}}{2}\right) = \pm \sqrt{\frac{1 - \cos \frac{\pi }{4}}{2}} = \pm \sqrt{\frac{1 - \frac{\sqrt{2}}{2}}{2}} = \pm \sqrt{\frac{\frac{2 - \sqrt{2}}{2}}{2}} \] \[ = \pm \sqrt{\frac{2 - \sqrt{2}}{4}} = \pm \frac{\sqrt{2 - \sqrt{2}}}{2}. \] Since \( \frac{\pi }{8} = \frac{{180}^{ \circ }}{8} = {22.5}^{ \circ } \) is in the first quadrant, the sine is positive, so that \( \sin \left( \frac{\pi }{8}\right) = \frac{\sqrt{2 - \sqrt{2}}}{2} \). b) Note that \( \frac{9\pi }{8} = \frac{\frac{9\pi }{4}}{2} \). So we use \( \alpha = \frac{9\pi }{4} \). Now, \( \frac{9\pi }{8} = \frac{9 \cdot {180}^{ \circ }}{8} = {202.5}^{ \circ } \) is in the third quadrant, so that the cosine is negative. We have: \[ \cos \left( \frac{9\pi }{8}\right) = \cos \left( \frac{\frac{9\pi }{4}}{2}\right) = - \sqrt{\frac{1 + \cos \frac{9\pi }{4}}{2}}. \] Now, \( \cos \left( \frac{9\pi }{4}\right) = \cos \left( \frac{{8\pi } + \pi }{4}\right) = \cos \left( {{2\pi } + \frac{\pi }{4}}\right) = \cos \left( \frac{\pi }{4}\right) = \frac{\sqrt{2}}{2} \), so that \[ \cos \left( \frac{9\pi }{8}\right) = - \sqrt{\frac{1 + \frac{\sqrt{2}}{2}}{2}} = - \sqrt{\frac{\frac{2 + \sqrt{2}}{2}}{2}} = - \sqrt{\frac{2 + \sqrt{2}}{4}} = - \frac{\sqrt{2 + \sqrt{2}}}{2}. \]
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Yes
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Example 18.7. Find the trigonometric functions of \( {2\alpha } \) when \( \alpha \) has the properties below.\n\na) \( \sin \left( \alpha \right) = \frac{3}{5} \), and \( \alpha \) is in quadrant II\n\nb) \( \tan \left( \alpha \right) = \frac{12}{5} \), and \( \alpha \) is in quadrant III
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Solution. a) From \( {\sin }^{2}\left( \alpha \right) + {\cos }^{2}\left( \alpha \right) = 1 \), we find that \( {\cos }^{2}\left( \alpha \right) = 1 - {\sin }^{2}\left( \alpha \right) \),\n\nand since \( \alpha \) is in the second quadrant, \( \cos \left( \alpha \right) \) is negative, so that\n\n\[ \cos \left( \alpha \right) = - \sqrt{1 - {\sin }^{2}\left( \alpha \right) } = - \sqrt{1 - {\left( \frac{3}{5}\right) }^{2}} = - \sqrt{1 - \frac{9}{25}} \]\n\n\[ = - \sqrt{\frac{{25} - 9}{25}} = - \sqrt{\frac{16}{25}} = - \frac{4}{5} \]\n\nand\n\n\[ \tan \left( \alpha \right) = \frac{\sin \alpha }{\cos \alpha } = \frac{\frac{3}{5}}{\frac{-4}{5}} = \frac{3}{5} \cdot \frac{5}{-4} = - \frac{3}{4} \]\n\nFrom this we can calculate the solution by plugging these values into the double angle formulas.\n\n\[ \sin \left( {2\alpha }\right) = 2\sin \alpha \cos \alpha = 2 \cdot \frac{3}{5} \cdot \frac{\left( -4\right) }{5} = \frac{-{24}}{25} \]\n\n\[ \cos \left( {2\alpha }\right) = {\cos }^{2}\left( \alpha \right) - {\sin }^{2}\left( \alpha \right) = {\left( \frac{-4}{5}\right) }^{2} - {\left( \frac{3}{5}\right) }^{2} = \frac{16}{25} - \frac{9}{25} = \frac{7}{25} \]\n\n\[ \tan \left( {2\alpha }\right) = \frac{2\tan \alpha }{1 - {\tan }^{2}\alpha } = \frac{2 \cdot \left( \frac{-3}{4}\right) }{1 - {\left( \frac{-3}{4}\right) }^{2}} = \frac{\frac{-3}{2}}{1 - \frac{9}{16}} = \frac{\frac{-3}{2}}{\frac{{16} - 9}{16}} = \frac{-3}{2} \cdot \frac{16}{7} = \frac{-{24}}{7} \]
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Yes
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Recall the exact values of the tangent function from section 17.1
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<table><tr><td>\\( x \\)</td><td>\\( 0 = {0}^{ \\circ } \\)</td><td>\\( \\frac{\\pi }{6} = {30}^{ \\circ } \\)</td><td>\\( \\frac{\\pi }{4} = {45}^{ \\circ } \\)</td><td>\\( \\frac{\\pi }{3} = {60}^{ \\circ } \\)</td><td>\\( \\frac{\\pi }{2} = {90}^{ \\circ } \\)</td></tr><tr><td>\\( \\tan \\left( x\\right) \\)</td><td>0</td><td>\\( \\frac{\\sqrt{3}}{3} \\)</td><td>1</td><td>\\( \\sqrt{3} \\)</td><td>undef.</td></tr></table>\n\nFrom this, we can deduce function values by reversing inputs and outputs, such as:\n\n\\[ \n\\tan \\left( \\frac{\\pi }{6}\\right) = \\frac{\\sqrt{3}}{3} \\Rightarrow {\\tan }^{-1}\\left( \\frac{\\sqrt{3}}{3}\\right) = \\frac{\\pi }{6} \n\\]\n\n\\[ \n\\tan \\left( \\frac{\\pi }{4}\\right) = 1 \\Rightarrow {\\tan }^{-1}\\left( 1\\right) = \\frac{\\pi }{4} \n\\]\n\nAlso, since \\( {\\tan }^{-1}\\left( {-x}\\right) = - {\\tan }^{-1}\\left( x\\right) \\) , we obtain the inverse tangent of negative numbers.\n\n\\[ \n{\\tan }^{-1}\\left( {-\\sqrt{3}}\\right) = - {\\tan }^{-1}\\left( \\sqrt{3}\\right) = - \\frac{\\pi }{3} \n\\]\n\n\\[ \n{\\tan }^{-1}\\left( {-1}\\right) = - {\\tan }^{-1}\\left( 1\\right) = - \\frac{\\pi }{4} \n\\]
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Yes
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Example 19.7. We first recall the known values of the sine.
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<table><tr><td>\\( x \\)</td><td>\\( 0 = {0}^{ \\circ } \\)</td><td>\\( \\frac{\\pi }{6} = {30}^{ \\circ } \\)</td><td>\\( \\frac{\\pi }{4} = {45}^{ \\circ } \\)</td><td>\\( \\frac{\\pi }{3} = {60}^{ \\circ } \\)</td><td>\\( \\frac{\\pi }{2} = {90}^{ \\circ } \\)</td></tr><tr><td>\\( \\sin \\left( x\\right) \\)</td><td>0</td><td>\\( \\frac{1}{2} \\)</td><td>\\( \\frac{\\sqrt{2}}{2} \\)</td><td>\\( \\frac{\\sqrt{3}}{2} \\)</td><td>1</td></tr></table>\n\nThese values together with the fact that the inverse sine is odd, that is \\( {\\sin }^{-1}\\left( {-x}\\right) = - {\\sin }^{-1}\\left( x\\right) \\), provides us with examples of its function values.\n\n\\[ \n{\\sin }^{-1}\\left( \\frac{\\sqrt{2}}{2}\\right) = \\frac{\\pi }{4},\\;{\\sin }^{-1}\\left( 1\\right) = \\frac{\\pi }{2}, \n\\]\n\n\\[ \n{\\sin }^{-1}\\left( 0\\right) = 0,\\;{\\sin }^{-1}\\left( \\frac{-1}{2}\\right) = - {\\sin }^{-1}\\left( \\frac{1}{2}\\right) = - \\frac{\\pi }{6}. \n\\]\n\nNote, that the domain of \\( y = {\\sin }^{-1}\\left( x\\right) \\) is \\( D = \\left\\lbrack {-1,1}\\right\\rbrack \\), so that input numbers that are not in this interval give undefined outputs of the inverse sine:\n\n\\[ \n{\\sin }^{-1}\\left( 3\\right) \\text{is undefined} \n\\]
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Yes
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Example 19.10. We first recall the known values of the cosine.
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<table><tr><td>\\( x \\)</td><td>\\( 0 = {0}^{ \\circ } \\)</td><td>\\( \\frac{\\pi }{6} = {30}^{ \\circ } \\)</td><td>\\( \\frac{\\pi }{4} = {45}^{ \\circ } \\)</td><td>\\( \\frac{\\pi }{3} = {60}^{ \\circ } \\)</td><td>\\( \\frac{\\pi }{2} = {90}^{ \\circ } \\)</td></tr><tr><td>\\( \\cos \\left( x\\right) \\)</td><td>1</td><td>\\( \\frac{\\sqrt{3}}{2} \\)</td><td>\\( \\frac{\\sqrt{2}}{2} \\)</td><td>\\( \\frac{1}{2} \\)</td><td>0</td></tr></table>\n\nHere are some examples for function values of the inverse cosine.\n\n\\[ \n{\\cos }^{-1}\\left( \\frac{\\sqrt{3}}{2}\\right) = \\frac{\\pi }{6},\\;{\\cos }^{-1}\\left( 1\\right) = 0,\\;{\\cos }^{-1}\\left( 0\\right) = \\frac{\\pi }{2}.\n\\]\n\nNegative inputs to the arccosine can be calculated with equation (19.3), that is \\( {\\cos }^{-1}\\left( {-x}\\right) = \\pi - {\\cos }^{-1}\\left( x\\right) \\), or by going back to the unit circle definition.\n\n\\[ \n{\\cos }^{-1}\\left( {-\\frac{1}{2}}\\right) = \\pi - {\\cos }^{-1}\\left( \\frac{1}{2}\\right) = \\pi - \\frac{\\pi }{3} = \\frac{{3\\pi } - \\pi }{3} = \\frac{2\\pi }{3},\n\\]\n\n\n\\[ \n{\\cos }^{-1}\\left( {-1}\\right) = \\pi - {\\cos }^{-1}\\left( 1\\right) = \\pi - 0 = \\pi .\n\\]\n\nFurthermore, the domain of \\( y = {\\cos }^{-1}\\left( x\\right) \\) is \\( D = \\left\\lbrack {-1,1}\\right\\rbrack \\) , so that input numbers not in this interval give undefined outputs of the inverse cosine.\n\n\\[ \n{\\cos }^{-1}\\left( {17}\\right) \\text{is undefined}\n\\]
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Yes
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Solve for \( x : \;\tan \left( x\right) = \sqrt{3} \)
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Solution. There is an obvious solution given by \( x = {\tan }^{-1}\left( \sqrt{3}\right) = \frac{\pi }{3} \), as we studied in the last section. However, we can look for all solutions of \( \tan \left( x\right) = \sqrt{3} \) by studying the graph of the tangent function, that is, by finding all points where the graph of the \( y = \tan \left( x\right) \) intersects with the horizontal line \( y = \sqrt{3} \). Since the function \( y = \tan \left( x\right) \) is periodic with period \( \pi \), we see that the other solutions of \( \tan \left( x\right) = \sqrt{3} \) besides \( x = \frac{\pi }{3} \) are\n\n\[ \frac{\pi }{3} + \pi ,\;\frac{\pi }{3} + {2\pi },\;\frac{\pi }{3} + {3\pi },\ldots ,\;\text{ and }\;\frac{\pi }{3} - \pi ,\;\frac{\pi }{3} - {2\pi },\;\frac{\pi }{3} - {3\pi },\ldots \]\n\nIn general, we write the solution as\n\n\[ x = \frac{\pi }{3} + n \cdot \pi ,\;\text{ where }n = 0, \pm 1, \pm 2, \pm 3,\ldots \]\n\nThe graph also shows that these are indeed all solutions of \( \tan \left( x\right) = \sqrt{3} \) .
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Yes
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Solve for \( x \) :\n\n\[ \text{a)}\tan \left( x\right) = 1\text{, b)}\tan \left( x\right) = - 1\text{, c)}\tan \left( x\right) = {5.1}\text{, d)}\tan \left( x\right) = - {3.7} \]
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Solution. a) First, we find \( {\tan }^{-1}\left( 1\right) = \frac{\pi }{4} \) . The general solution is thus:\n\n\[ x = \frac{\pi }{4} + n \cdot \pi \;\text{ where }n = 0, \pm 1, \pm 2, \pm 3,\ldots \]\n\n\nb) First, we need to find \( {\tan }^{-1}\left( {-1}\right) \) . Recall from equation (19.1) that \( {\tan }^{-1}\left( {-c}\right) = - {\tan }^{-1}\left( c\right) \), and recall further that \( {\tan }^{-1}\left( 1\right) = \frac{\pi }{4} \) . With this we\n\nhave\n\[ {\tan }^{-1}\left( {-1}\right) = - {\tan }^{-1}\left( 1\right) = - \frac{\pi }{4} \]\n\nThe general solution of \( \tan \left( x\right) = - 1 \) is therefore,\n\n\[ x = - \frac{\pi }{4} + n \cdot \pi ,\;\text{ where }n = 0, \pm 1, \pm 2,\ldots \]\n\nFor parts (c) and (d), we do not have an exact solution, so that the solution can only be approximated with the calculator.\n\nc) \( x = {\tan }^{-1}\left( {5.1}\right) + {n\pi }\; \approx {1.377} + {n\pi },\; \) where \( n = 0, \pm 1, \pm 2,\ldots \]\n\nd) \( x = {\tan }^{-1}\left( {-{3.7}}\right) + {n\pi }\; \approx - {1.307} + {n\pi },\; \) where \( n = 0, \pm 1, \pm 2,\ldots \]
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Yes
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Solve for \( x : \;\cos \left( x\right) = \frac{1}{2} \)
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Solution. We have the obvious solution to the equation \( x = {\cos }^{-1}\left( \frac{1}{2}\right) = \frac{\pi }{3} \) . However, since \( \cos \left( {-x}\right) = \cos \left( x\right) \), there is another solution given by taking \( x = - \frac{\pi }{3} \) :\n\n\[ \cos \left( {-\frac{\pi }{3}}\right) = \cos \left( \frac{\pi }{3}\right) = \frac{1}{2} \]\n\nMoreover, the \( y = \cos \left( x\right) \) function is periodic with period \( {2\pi } \), that is, we have \( \cos \left( {x + {2\pi }}\right) = \cos \left( x\right) \) . Thus, all of the following numbers are solutions of the equation \( \cos \left( x\right) = \frac{1}{2} \) :\n\n\[ \ldots ,\;\frac{\pi }{3} - {4\pi },\;\frac{\pi }{3} - {2\pi },\;\frac{\pi }{3},\;\frac{\pi }{3} + {2\pi },\;\frac{\pi }{3} + {4\pi },\;\ldots ,\]\n\n\[ \text{and:}\ldots ,\; - \frac{\pi }{3} - {4\pi },\; - \frac{\pi }{3} - {2\pi },\; - \frac{\pi }{3},\; - \frac{\pi }{3} + {2\pi },\; - \frac{\pi }{3} + {4\pi },\;\ldots \]\n\nFrom the graph we see that there are only two solutions of \( \cos \left( x\right) = \frac{1}{2} \) within one period. Thus, the above list constitutes all solutions of the equation. With this observation, we may write the general solution as:\n\n\[ x = \frac{\pi }{3} + {2n} \cdot \pi ,\text{ or }x = - \frac{\pi }{3} + {2n} \cdot \pi ,\;\text{ where }n = 0, \pm 1, \pm 2, \pm 3,\ldots \]\n\nIn short, we write this as: \( x = \pm \frac{\pi }{3} + {2n} \cdot \pi \) with \( n = 0, \pm 1, \pm 2, \pm 3,\ldots \) .
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Yes
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Solve for \( x \). \[ \text{a)}\cos \left( x\right) = - \frac{\sqrt{2}}{2} \]
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a) First, we need to find \( {\cos }^{-1}\left( {-\frac{\sqrt{2}}{2}}\right) \). From equation (19.3) we know that \( {\cos }^{-1}\left( {-c}\right) = \pi - {\cos }^{-1}\left( c\right) \), so that: \[ {\cos }^{-1}\left( {-\frac{\sqrt{2}}{2}}\right) = \pi - {\cos }^{-1}\left( \frac{\sqrt{2}}{2}\right) = \pi - \frac{\pi }{4} = \frac{{4\pi } - \pi }{4} = \frac{3\pi }{4} \] The solution is therefore, \[ x = \pm \frac{3\pi }{4} + {2n\pi },\;\text{ where }n = 0, \pm 1, \pm 2, \pm 3,\ldots \]
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Yes
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Solve for \( x : \;\sin \left( x\right) = \frac{\sqrt{2}}{2} \)
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Solution. First, we can find one obvious solution \( x = {\sin }^{-1}\left( \frac{\sqrt{2}}{2}\right) = \frac{\pi }{4} \) . Furthermore, from the top right equation in (17.3), we have that \( \sin \left( {\pi - x}\right) = \sin \left( x\right) \) , so that another solution is given by \( \pi - \frac{\pi }{4} \) :\n\n\[ \sin \left( {\pi - \frac{\pi }{4}}\right) = \sin \left( \frac{\pi }{4}\right) = \frac{\sqrt{2}}{2} \]\n\n(This can also be seen by going back to the unit circle definition.) These are all solutions within one period, as can be checked from the graph above. The function \( y = \sin \left( x\right) \) is periodic with period \( {2\pi } \), so that adding \( {2n} \cdot \pi \) for any \( n = 0, \pm 1, \pm 2,\ldots \) gives all solutions of \( \sin \left( x\right) = \frac{\sqrt{2}}{2} \) . This means that the general solution is:\n\n\[ x = \frac{\pi }{4} + {2n} \cdot \pi ,\text{ or }x = \left( {\pi - \frac{\pi }{4}}\right) + {2n} \cdot \pi ,\;\text{ for }n = 0, \pm 1, \pm 2, \pm 3,\ldots \]\n\nWe rewrite these solutions to obtain one single formula for the solutions. Note, that \( \pi - \frac{\pi }{4} + {2n} \cdot \pi = - \frac{\pi }{4} + \left( {{2n} + 1}\right) \cdot \pi \) . Therefore, all solutions are of the form\n\n\[ x = \pm \frac{\pi }{4} + k \cdot \pi \]\n\nwhere, for even \( k = {2n} \), the sign in front of \( \frac{\pi }{4} \) is \
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Yes
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Solve for \( x \). a) \(\sin \left( x\right) = \frac{1}{2}\)
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a) First, we calculate \( {\sin }^{-1}\left( \frac{1}{2}\right) = \frac{\pi }{6} \). The general solution is therefore,\n\n\[ x = {\left( -1\right) }^{n} \cdot \frac{\pi }{6} + n \cdot \pi ,\;\text{ where }n = 0, \pm 1, \pm 2, \pm 3,\ldots \]
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Yes
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Find the general solution of the equation, and state at least 5 distinct solutions.
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\[ \text{a)}\sin \left( x\right) = - \frac{1}{2}\;\text{b)}\cos \left( x\right) = - \frac{\sqrt{3}}{2} \] Solution. a) We already calculated the general solution in example 20.9 (b). It is \[ x = {\left( -1\right) }^{n + 1} \cdot \frac{\pi }{6} + n \cdot \pi ,\;\text{ where }n = 0, \pm 1, \pm 2, \pm 3,\ldots \] We simplify the solutions for \( n = 0,1, - 1,2, - 2 \) . \[ n = 0 : x = {\left( -1\right) }^{0 + 1} \cdot \frac{\pi }{6} + 0 \cdot \pi = - \frac{\pi }{6} \] \[ n = 1 : x = {\left( -1\right) }^{1 + 1} \cdot \frac{\pi }{6} + 1 \cdot \pi = \frac{\pi }{6} + \pi = \frac{\pi + {6\pi }}{6} = \frac{7\pi }{6} \] \[ n = - 1 : x = {\left( -1\right) }^{-1 + 1} \cdot \frac{\pi }{6} + \left( {-1}\right) \cdot \pi = \frac{\pi }{6} - \pi = \frac{\pi - {6\pi }}{6} = \frac{-{5\pi }}{6} \] \[ n = 2 : x = {\left( -1\right) }^{2 + 1} \cdot \frac{\pi }{6} + 2 \cdot \pi = - \frac{\pi }{6} + {2\pi } = \frac{-\pi + {12\pi }}{6} = \frac{11\pi }{6} \] \[ n = - 2 : x = {\left( -1\right) }^{-2 + 1} \cdot \frac{\pi }{6} + \left( {-2}\right) \cdot \pi = - \frac{\pi }{6} - {2\pi } = \frac{-\pi - {12\pi }}{6} = \frac{-{13\pi }}{6} \]
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Yes
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Example 20.11. Solve for \( x \). \[ \text{a)}2\sin \left( x\right) - 1 = 0\;\text{b)}\sec \left( x\right) = - \sqrt{2}\;\text{c)}7\cot \left( x\right) + 3 = 0 \]
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Solution. a) Solving for \( \sin \left( x\right) \), we get \[ 2\sin \left( x\right) - 1 = 0\overset{\left( +1\right) }{ \Rightarrow }2\sin \left( x\right) = 1\overset{\left( \div 2\right) }{ \Rightarrow }\sin \left( x\right) = \frac{1}{2} \] One solution of \( \sin \left( x\right) = \frac{1}{2} \) is \( {\sin }^{-1}\left( \frac{1}{2}\right) = \frac{\pi }{6} \). The general solution is \[ x = {\left( -1\right) }^{n} \cdot \frac{\pi }{6} + {n\pi },\;\text{ where }n = 0, \pm 1, \pm 2,\ldots \] b) Recall that \( \sec \left( x\right) = \frac{1}{\cos \left( x\right) } \). Therefore, \[ \sec \left( x\right) = - \sqrt{2} \Rightarrow \frac{1}{\cos \left( x\right) } = - \sqrt{2}\;\overset{\text{ (reciprocal) }}{ \Rightarrow }\;\cos \left( x\right) = - \frac{1}{\sqrt{2}} = - \frac{\sqrt{2}}{2} \] A special solution of \( \cos \left( x\right) = - \frac{\sqrt{2}}{2} \) is \[ {\cos }^{-1}\left( {-\frac{\sqrt{2}}{2}}\right) = \pi - {\cos }^{-1}\left( \frac{\sqrt{2}}{2}\right) = \pi - \frac{\pi }{4} = \frac{{4\pi } - \pi }{4} = \frac{3\pi }{4}. \] The general solution is \[ x = \pm \frac{3\pi }{4} + {2n\pi },\;\text{ where }n = 0, \pm 1, \pm 2,\ldots \] c) Recall that \( \cot \left( x\right) = \frac{1}{\tan \left( x\right) } \). So \[ 7\cot (x) + 3 = 0\;\overset{( - 3)}{\;\Longrightarrow \;}\;7\cot (x) = - 3\;\overset{( \div 7)}{\;\Longrightarrow \;}\;\cot (x) = - \frac{3}{7} \] \[ \Rightarrow \frac{1}{\tan \left( x\right) } = - \frac{3}{7}\;\overset{\text{ (reciprocal) }}{ \Rightarrow }\;\tan \left( x\right) = - \frac{7}{3} \] The solution is \[ x = {\tan }^{-1}\left( {-\frac{7}{3}}\right) + {n\pi } \approx - {1.166} + {n\pi },\;\text{ where }n = 0, \pm 1, \pm 2,\ldots \]
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Yes
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Example 20.13. Solve the equation with the calculator. Approximate the solution to the nearest thousandth.\n\n\\[ \n\\text{a)}2\\sin \\left( x\\right) = 4\\cos \\left( x\\right) + 3\\;\\text{b)}5\\cos \\left( {2x}\\right) = \\tan \\left( x\\right) \n\\]
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Solution. a) We rewrite the equation as \\( 2\\sin \\left( x\\right) - 4\\cos \\left( x\\right) - 3 = 0 \\), and use the calculator to find the graph of the function \\( f\\left( x\\right) = 2\\sin \\left( x\\right) - 4\\cos \\left( x\\right) - 3 \\) . The zeros of the function \\( f \\) are the solutions of the initial equation. The graph that we obtain is displayed below.\n\n\n\nThe graph indicates that the function \\( f\\left( x\\right) = 2\\sin \\left( x\\right) - 4\\cos \\left( x\\right) - 3 \\) is periodic. This can be confirmed by observing that both \\( \\sin \\left( x\\right) \\) and \\( \\cos \\left( x\\right) \\) are periodic with period \\( {2\\pi } \\), and thus also \\( f\\left( x\\right) \\) .\n\n\\[ \nf\\left( {x + {2\\pi }}\\right) = 2\\sin \\left( {x + {2\\pi }}\\right) - 4\\cos \\left( {x + {2\\pi }}\\right) - 3 = 2\\sin \\left( x\\right) - 4\\cos \\left( x\\right) - 3 = f\\left( x\\right) \n\\]\n\nThe solution of \\( f\\left( x\\right) = 0 \\) can be obtained by finding the \
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No
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Example 21.3. Perform the operation.\n\n\[ \text{a)}\left( {2 - {3i}}\right) + \left( {-6 + {4i}}\right) \]
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Solution. a) Adding real and imaginary parts, respectively, gives,\n\n\[ \left( {2 - {3i}}\right) + \left( {-6 + {4i}}\right) = 2 - {3i} - 6 + {4i} = - 4 + i. \]
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Yes
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Example 21.5. Find the absolute value of the complex numbers below.
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Solution. The absolute values are calculated as follows.\n\n\[ \text{a)}\left| {5 - {3i}}\right| = \sqrt{{5}^{2} + {\left( -3\right) }^{2}} = \sqrt{{25} + 9} = \sqrt{34} \]\n\nb) \( \;\left| {-8 + {6i}}\right| = \sqrt{{\left( -8\right) }^{2} + {6}^{2}} = \sqrt{{64} + {36}} = \sqrt{100} = {10} \)\n\nc) \( \;\left| {7i}\right| = \sqrt{{0}^{2} + {\left( 7\right) }^{2}} = \sqrt{0 + {49}} = 7 \)
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Yes
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