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Example 2.3.5. Graph \( f\left( x\right) = \left| {{x}^{2} - x - 6}\right| \) .
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Solution. Using the definition of absolute value, Definition 2.4, we have\n\n\[ f\left( x\right) = \left\{ \begin{array}{rrr} - \left( {{x}^{2} - x - 6}\right) , & \text{ if } & {x}^{2} - x - 6 < 0 \\ {x}^{2} - x - 6, & \text{ if } & {x}^{2} - x - 6 \geq 0 \end{array}\right. \]\n\nThe trouble is that we have yet to develop any analytic techniques to solve nonlinear inequalities such as \( {x}^{2} - x - 6 < 0 \) . You won’t have to wait long; this is one of the main topics of Section 2.4.\n\n--- \n\nNevertheless, we can attack this problem graphically. To that end, we graph \( y = g\left( x\right) = {x}^{2} - x - 6 \) using the intercepts and the vertex. To find the \( x \) -intercepts, we solve \( {x}^{2} - x - 6 = 0 \) . Factoring gives \( \left( {x - 3}\right) \left( {x + 2}\right) = 0 \) so \( x = - 2 \) or \( x = 3 \) . Hence, \( \left( {-2,0}\right) \) and \( \left( {3,0}\right) \) are \( x \) -intercepts. The \( y \) -intercept \( \left( {0, - 6}\right) \) is found by setting \( x = 0 \) . To plot the vertex, we find \( x = - \frac{b}{2a} = - \frac{-1}{2\left( 1\right) } = \frac{1}{2} \), and \( y = {\left( \frac{1}{2}\right) }^{2} - \left( \frac{1}{2}\right) - 6 = - \frac{25}{4} = - {6.25} \) . Plotting, we get the parabola seen below on the left. To obtain points on the graph of \( y = f\left( x\right) = \left| {{x}^{2} - x - 6}\right| \), we can take points on the graph of \( g\left( x\right) = {x}^{2} - x - 6 \) and apply the absolute value to each of the \( y \) values on the parabola. We see from the graph of \( g \) that for \( x \leq - 2 \) or \( x \geq 3 \), the \( y \) values on the parabola are greater than or equal to zero (since the graph is on or above the \( x \) -axis), so the absolute value leaves these portions of the graph alone. For \( x \) between -2 and 3, however, the \( y \) values on the parabola are negative. For example, the point \( \left( {0, - 6}\right) \) on \( y = {x}^{2} - x - 6 \) would result in the point \( \left( {0,\left| {-6}\right| }\right) = \left( {0, - \left( {-6}\right) }\right) = \left( {0,6}\right) \) on the graph of \( f\left( x\right) = \left| {{x}^{2} - x - 6}\right| \) . Proceeding in this manner for all points with \( x \) -coordinates between -2 and 3 results in the graph seen below on the right. 
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Yes
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1. Solve \( f\left( x\right) = g\left( x\right) \) .
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1. To solve \( f\left( x\right) = g\left( x\right) \), we replace \( f\left( x\right) \) with \( {2x} - 1 \) and \( g\left( x\right) \) with 5 to get \( {2x} - 1 = 5 \) . Solving for \( x \), we get \( x = 3 \) .
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Yes
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1. Solve \( f\left( x\right) = g\left( x\right) \) .
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1. To solve \( f\left( x\right) = g\left( x\right) \), we look for where the graphs of \( f \) and \( g \) intersect. These appear to be at the points \( \left( {-1,2}\right) \) and \( \left( {1,2}\right) \), so our solutions to \( f\left( x\right) = g\left( x\right) \) are \( x = - 1 \) and \( x = 1 \) .
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Yes
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Theorem 2.4. Inequalities Involving the Absolute Value: Let \( c \) be a real number.\n\n- For \( c > 0,\left| x\right| < c \) is equivalent to \( - c < x < c \) .\n\n- For \( c > 0,\left| x\right| \leq c \) is equivalent to \( - c \leq x \leq c \) .\n\n- For \( c \leq 0,\left| x\right| < c \) has no solution, and for \( c < 0,\left| x\right| \leq c \) has no solution.\n\n- For \( c \geq 0,\left| x\right| > c \) is equivalent to \( x < - c \) or \( x > c \) .\n\n- For \( c \geq 0,\left| x\right| \geq c \) is equivalent to \( x \leq - c \) or \( x \geq c \) .\n\n- For \( c < 0,\left| x\right| > c \) and \( \left| x\right| \geq c \) are true for all real numbers.
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As with Theorem 2.1 in Section 2.2, we could argue Theorem 2.4 using cases. However, in light of what we have developed in this section, we can understand these statements graphically. For instance, if \( c > 0 \), the graph of \( y = c \) is a horizontal line which lies above the \( x \) -axis through \( \left( {0, c}\right) \) . To solve \( \left| x\right| < c \), we are looking for the \( x \) values where the graph of \( y = \left| x\right| \) is below the graph of \( y = c \) . We know that the graphs intersect when \( \left| x\right| = c \), which, from Section 2.2, we know happens when \( x = c \) or \( x = - c \) . Graphing, we get\n\n\n\nWe see that the graph of \( y = \left| x\right| \) is below \( y = c \) for \( x \) between \( - c \) and \( c \), and hence we get \( \left| x\right| < c \) is equivalent to \( - c < x < c \) . The other properties in Theorem 2.4 can be shown similarly.
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Yes
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1. \( \left| {x - 1}\right| \geq 3 \)
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From Theorem 2.4, \( \left| {x - 1}\right| \geq 3 \) is equivalent to \( x - 1 \leq - 3 \) or \( x - 1 \geq 3 \) . Solving, we get \( x \leq - 2 \) or \( x \geq 4 \), which, in interval notation is \( \left( {-\infty , - 2\rbrack \cup \lbrack 4,\infty }\right) \) . Graphically, we have\n\n\n\nWe see that the graph of \( y = \left| {x - 1}\right| \) is above the horizontal line \( y = 3 \) for \( x < - 2 \) and \( x > 4 \) hence this is where \( \left| {x - 1}\right| > 3 \) . The two graphs intersect when \( x = - 2 \) and \( x = 4 \), so we have graphical confirmation of our analytic solution.
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Yes
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1. \( 2{x}^{2} \leq 3 - x \)
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To solve \( 2{x}^{2} \leq 3 - x \), we first get 0 on one side of the inequality which yields \( 2{x}^{2} + x - 3 \leq 0 \) . We find the zeros of \( f\left( x\right) = 2{x}^{2} + x - 3 \) by solving \( 2{x}^{2} + x - 3 = 0 \) for \( x \) . Factoring gives \( \left( {{2x} + 3}\right) \left( {x - 1}\right) = 0 \), so \( x = - \frac{3}{2} \) or \( x = 1 \) . We place these values on the number line with 0 above them and choose test values in the intervals \( \left( {-\infty , - \frac{3}{2}}\right) ,\left( {-\frac{3}{2},1}\right) \) and \( \left( {1,\infty }\right) \) . For the interval \( \left( {-\infty , - \frac{3}{2}}\right) \), we choose \( {}^{3}x = - 2 \) ; for \( \left( {-\frac{3}{2},1}\right) \), we pick \( x = 0 \) ; and for \( \left( {1,\infty }\right), x = 2 \) . Evaluating the function at the three test values gives us \( f\left( {-2}\right) = 3 > 0 \), so we place \( \left( +\right) \) above \( \left( {-\infty , - \frac{3}{2}}\right) ;f\left( 0\right) = - 3 < 0 \), so \( \left( -\right) \) goes above the interval \( \left( {-\frac{3}{2},1}\right) \) ; and, \( f\left( 2\right) = 7 \) , which means \( \left( +\right) \) is placed above \( \left( {1,\infty }\right) \) . Since we are solving \( 2{x}^{2} + x - 3 \leq 0 \), we look for solutions to \( 2{x}^{2} + x - 3 < 0 \) as well as solutions for \( 2{x}^{2} + x - 3 = 0 \) . For \( 2{x}^{2} + x - 3 < 0 \), we need the intervals which we have a \( \left( -\right) \) . Checking the sign diagram, we see this is \( \left( {-\frac{3}{2},1}\right) \) . We know \( 2{x}^{2} + x - 3 = 0 \) when \( x = - \frac{3}{2} \) and \( x = 1 \), so our final answer is \( \left\lbrack {-\frac{3}{2},1}\right\rbrack \) .
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Yes
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Example 2.4.5. The area \( A \) (in square inches) of a square piece of particle board which measures \( x \) inches on each side is \( A\left( x\right) = {x}^{2} \) . Suppose a manufacturer needs to produce a 24 inch by 24 inch square piece of particle board as part of a home office desk kit. How close does the side of the piece of particle board need to be cut to 24 inches to guarantee that the area of the piece is within a tolerance of 0.25 square inches of the target area of 576 square inches?
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Solution. Mathematically, we express the desire for the area \( A\left( x\right) \) to be within 0.25 square inches of 576 as \( \left| {A - {576}}\right| \leq {0.25} \) . Since \( A\left( x\right) = {x}^{2} \), we get \( \left| {{x}^{2} - {576}}\right| \leq {0.25} \), which is equivalent to \( - {0.25} \leq {x}^{2} - {576} \leq {0.25} \) . One way to proceed at this point is to solve the two inequalities \( - {0.25} \leq {x}^{2} - {576} \) and \( {x}^{2} - {576} \leq {0.25} \) individually using sign diagrams and then taking the intersection of the solution sets. While this way will (eventually) lead to the correct answer, we take this opportunity to showcase the increasing property of the square root: if \( 0 \leq a \leq b \), then \( \sqrt{a} \leq \sqrt{b} \) . To use this property, we proceed as follows\n\n\[ - {0.25} \leq {x}^{2} - {576} \leq {0.25} \]\n\n\[ {575.75} \leq \;{x}^{2}\; \leq {576.25}\;\text{(add 576 across the inequalities.)} \]\n\n\[ \sqrt{575.75} \leq \;\left| x\right| \; \leq \sqrt{576.25}\;\left( {\sqrt{{x}^{2}} = \left| x\right| }\right) \]\n\nBy Theorem 2.4, we find the solution to \( \sqrt{575.75} \leq \left| x\right| \) to be \( \left( {-\infty , - \sqrt{575.75}\rbrack \cup \lbrack \sqrt{575.75},\infty }\right) \) and the solution to \( \left| x\right| \leq \sqrt{576.25} \) to be \( \left\lbrack {-\sqrt{576.25},\sqrt{576.25}}\right\rbrack \) . To solve \( \sqrt{575.75} \leq \left| x\right| \leq \sqrt{576.25} \), we intersect these two sets to get \( \left\lbrack {-\sqrt{576.25}, - \sqrt{575.75}}\right\rbrack \cup \left\lbrack {\sqrt{575.75},\sqrt{576.25}}\right\rbrack \) . Since \( x \) represents a length, we discard the negative answers and get \( \left\lbrack {\sqrt{575.75},\sqrt{576.25}}\right\rbrack \) . This means that the side of the piece of particle board must be cut between \( \sqrt{575.75} \approx {23.995} \) and \( \sqrt{576.25} \approx {24.005} \) inches, a tolerance of (approximately) 0.005 inches of the target length of 24 inches.
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Yes
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1. \( R = \{ \left( {x, y}\right) : y > \left| x\right| \} \)
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The relation \( R \) consists of all points \( \left( {x, y}\right) \) whose \( y \) -coordinate is greater than \( \left| x\right| \) . If we graph \( y = \left| x\right| \), then we want all of the points in the plane above the points on the graph. Dotting the graph of \( y = \left| x\right| \) as we have done before to indicate that the points on the graph itself are not in the relation, we get the shaded region below on the left.
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Yes
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2. Find the least squares regression line and comment on the goodness of fit.
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Performing a linear regression produces\n\n\n\nWe can tell both from the correlation coefficient as well as the graph that the regression line is a good fit to the data.
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No
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Example 2.5.2. Using the quadratic model for the temperature data above, predict the warmest temperature of the day. When will this occur?
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Solution. The maximum temperature will occur at the vertex of the parabola. Recalling the Vertex Formula, Equation 2.4, \( x = - \frac{b}{2a} \approx - \frac{9.464}{2\left( {-{0.321}}\right) } \approx {14.741} \) . This corresponds to roughly 2: 45 PM. To find the temperature, we substitute \( x = {14.741} \) into \( y = - {0.321}{x}^{2} + {9.464x} - {45.857} \) to get \( y \approx {23.899} \), or \( {23.899}^{ \circ }\mathrm{F} \) .
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Yes
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Example 3.1.1. Determine if the following functions are polynomials. Explain your reasoning.
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1. We note directly that the domain of \( g\left( x\right) = \frac{{x}^{3} + 4}{x} \) is \( x \neq 0 \) . By definition, a polynomial has all real numbers as its domain. Hence, \( g \) can’t be a polynomial.\n\n2. Even though \( p\left( x\right) = \frac{{x}^{3} + {4x}}{x} \) simplifies to \( p\left( x\right) = {x}^{2} + 4 \), which certainly looks like the form given in Definition 3.1, the domain of \( p \), which, as you may recall, we determine before we simplify, excludes 0 . Alas, \( p \) is not a polynomial function for the same reason \( g \) isn’t.\n\n3. After what happened with \( p \) in the previous part, you may be a little shy about simplifying \( q\left( x\right) = \frac{{x}^{3} + {4x}}{{x}^{2} + 4} \) to \( q\left( x\right) = x \), which certainly fits Definition 3.1. If we look at the domain of \( q \) before we simplified, we see that it is, indeed, all real numbers. A function which can be written in the form of Definition 3.1 whose domain is all real numbers is, in fact, a polynomial.\n\n4. We can rewrite \( f\left( x\right) = \sqrt[3]{x} \) as \( f\left( x\right) = {x}^{\frac{1}{3}} \) . Since \( \frac{1}{3} \) is not a natural number, \( f \) is not a polynomial.\n\n5. The function \( h\left( x\right) = \left| x\right| \) isn’t a polynomial, since it can’t be written as a combination of powers of \( x \) even though it can be written as a piecewise function involving polynomials. As we shall see in this section, graphs of polynomials possess a quality \( {}^{2} \) that the graph of \( h \) does not.\n\n6. There’s nothing in Definition 3.1 which prevents all the coefficients \( {a}_{n} \), etc., from being 0 . Hence, \( z\left( x\right) = 0 \), is an honest-to-goodness polynomial.
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Yes
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Example 3.1.2. Find the degree, leading term, leading coefficient and constant term of the following polynomial functions.\n\n1. \( f\left( x\right) = 4{x}^{5} - 3{x}^{2} + {2x} - 5 \) 2. \( g\left( x\right) = {12x} + {x}^{3} \)\n\n3. \( h\left( x\right) = \frac{4 - x}{5} \) 4. \( p\left( x\right) = {\left( 2x - 1\right) }^{3}\left( {x - 2}\right) \left( {{3x} + 2}\right) \)
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## Solution.\n\n1. There are no surprises with \( f\left( x\right) = 4{x}^{5} - 3{x}^{2} + {2x} - 5 \) . It is written in the form of Definition 3.2, and we see that the degree is 5, the leading term is \( 4{x}^{5} \), the leading coefficient is 4 and the constant term is -5 .\n\n2. The form given in Definition 3.2 has the highest power of \( x \) first. To that end, we re-write \( g\left( x\right) = {12x} + {x}^{3} = {x}^{3} + {12x} \), and see that the degree of \( g \) is 3, the leading term is \( {x}^{3} \), the leading coefficient is 1 and the constant term is 0 .\n\n3. We need to rewrite the formula for \( h \) so that it resembles the form given in Definition 3.2: \( h\left( x\right) = \frac{4 - x}{5} = \frac{4}{5} - \frac{x}{5} = - \frac{1}{5}x + \frac{4}{5} \) . The degree of \( h \) is 1, the leading term is \( - \frac{1}{5}x \), the leading coefficient is \( - \frac{1}{5} \) and the constant term is \( \frac{4}{5} \).\n\n4. It may seem that we have some work ahead of us to get \( p \) in the form of Definition 3.2. However, it is possible to glean the information requested about \( p \) without multiplying out the entire expression \( {\left( 2x - 1\right) }^{3}\left( {x - 2}\right) \left( {{3x} + 2}\right) \) . The leading term of \( p \) will be the term which has the highest power of \( x \) . The way to get this term is to multiply the terms with the highest power of \( x \) from each factor together - in other words, the leading term of \( p\left( x\right) \) is the product of the leading terms of the factors of \( p\left( x\right) \) . Hence, the leading term of \( p \) is \( {\left( 2x\right) }^{3}\left( x\right) \left( {3x}\right) = {24}{x}^{5} \) . This means that the degree of \( p \) is 5 and the leading coefficient is 24 . As for the constant term, we can perform a similar trick. The constant term is obtained by multiplying the constant terms from each of the factors \( {\left( -1\right) }^{3}\left( {-2}\right) \left( 2\right) = 4 \).
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Yes
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1. Find the volume \( V \) of the box as a function of \( x \) . Include an appropriate applied domain.
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## Solution.\n\n1. From Geometry, we know that Volume \( = \) width \( \times \) height \( \times \) depth. The key is to find each of these quantities in terms of \( x \) . From the figure, we see that the height of the box is \( x \) itself. The cardboard piece is initially 10 inches wide. Removing squares with a side length of \( x \) inches from each corner leaves \( {10} - {2x} \) inches for the width. \( {}^{5} \) As for the depth, the cardboard is initially 12 inches long, so after cutting out \( x \) inches from each side, we would have \( {12} - {2x} \) inches remaining. As a function \( {}^{6} \) of \( x \), the volume is\n\n\[ V\left( x\right) = x\left( {{10} - {2x}}\right) \left( {{12} - {2x}}\right) = 4{x}^{3} - {44}{x}^{2} + {120x} \]\n\nTo find a suitable applied domain, we note that to make a box at all we need \( x > 0 \) . Also the shorter of the two dimensions of the cardboard is 10 inches, and since we are removing \( {2x} \) inches from this dimension, we also require \( {10} - {2x} > 0 \) or \( x < 5 \) . Hence, our applied domain is \( 0 < x < 5 \) .
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Yes
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Use the Intermediate Value Theorem to establish that \( \sqrt{2} \) is a real number.
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Consider the polynomial function \( f\left( x\right) = {x}^{2} - 2 \) . Then \( f\left( 1\right) = - 1 \) and \( f\left( 3\right) = 7 \) . Since \( f\left( 1\right) \) and \( f\left( 3\right) \) have different signs, the Intermediate Value Theorem guarantees us a real number \( c \) between 1 and 3 with \( f\left( c\right) = 0 \) . If \( {c}^{2} - 2 = 0 \) then \( c = \pm \sqrt{2} \) . Since \( c \) is between 1 and \( 3, c \) is positive, so \( c = \sqrt{2} \).
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Yes
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Construct a sign diagram for \( f\left( x\right) = {x}^{3}{\left( x - 3\right) }^{2}\left( {x + 2}\right) \left( {{x}^{2} + 1}\right) \) . Use it to give a rough sketch of the graph of \( y = f\left( x\right) \) .
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Solution. First, we find the zeros of \( f \) by solving \( {x}^{3}{\left( x - 3\right) }^{2}\left( {x + 2}\right) \left( {{x}^{2} + 1}\right) = 0 \) . We get \( x = 0 \) , \( x = 3 \) and \( x = - 2 \) . (The equation \( {x}^{2} + 1 = 0 \) produces no real solutions.) These three points divide the real number line into four intervals: \( \left( {-\infty , - 2}\right) ,\left( {-2,0}\right) ,\left( {0,3}\right) \) and \( \left( {3,\infty }\right) \) . We select the test values \( x = - 3, x = - 1, x = 1 \) and \( x = 4 \) . We find \( f\left( {-3}\right) \) is \( \left( +\right), f\left( {-1}\right) \) is \( \left( -\right) \) and \( f\left( 1\right) \) is \( \left( +\right) \) as is \( f\left( 4\right) \) . Wherever \( f \) is \( \left( +\right) \), its graph is above the \( x \) -axis; wherever \( f \) is \( \left( -\right) \), its graph is below the \( x \) -axis. The \( x \) -intercepts of the graph of \( f \) are \( \left( {-2,0}\right) ,\left( {0,0}\right) \) and \( \left( {3,0}\right) \) . Knowing \( f \) is smooth and continuous allows us to sketch its graph.
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Yes
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Theorem 3.2. End Behavior for Polynomial Functions: The end behavior of a polynomial \( f\left( x\right) = {a}_{n}{x}^{n} + {a}_{n - 1}{x}^{n - 1} + \ldots + {a}_{2}{x}^{2} + {a}_{1}x + {a}_{0} \) with \( {a}_{n} \neq 0 \) matches the end behavior of \( y = {a}_{n}{x}^{n} \) .
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To see why Theorem 3.2 is true, let’s first look at a specific example. Consider \( f\left( x\right) = 4{x}^{3} - x + 5 \) . If we wish to examine end behavior, we look to see the behavior of \( f \) as \( x \rightarrow \pm \infty \) . Since we’re concerned with \( x \) ’s far down the \( x \) -axis, we are far away from \( x = 0 \) so can rewrite \( f\left( x\right) \) for these\n\nvalues of \( x \) as\n\[ f\left( x\right) = 4{x}^{3}\left( {1 - \frac{1}{4{x}^{2}} + \frac{5}{4{x}^{3}}}\right) \]\n\nAs \( x \) becomes unbounded (in either direction), the terms \( \frac{1}{4{x}^{2}} \) and \( \frac{5}{4{x}^{3}} \) become closer and closer to 0 , as the table below indicates.\n\n<table><thead><tr><th>\( x \)</th><th>\( \frac{1}{4{x}^{2}} \)</th><th>\( \frac{5}{4{x}^{3}} \)</th></tr></thead><tr><td>\( - {1000} \)</td><td>0.00000025</td><td>\( - {0.00000000125} \)</td></tr><tr><td>\( - {100} \)</td><td>0.000025</td><td>\( - {0.00000125} \)</td></tr><tr><td>\( - {10} \)</td><td>0.0025</td><td>\( - {0.00125} \)</td></tr><tr><td>10</td><td>0.0025</td><td>0.00125</td></tr><tr><td>100</td><td>0.000025</td><td>0.00000125</td></tr><tr><td>1000</td><td>0.00000025</td><td>0.00000000125</td></tr></table>\n\nIn other words, as \( x \rightarrow \pm \infty, f\left( x\right) \approx 4{x}^{3}\left( {1 - 0 + 0}\right) = 4{x}^{3} \), which is the leading term of \( f \) . The formal proof of Theorem 3.2 works in much the same way. Factoring out the leading term leaves\n\n\[ f\left( x\right) = {a}_{n}{x}^{n}\left( {1 + \frac{{a}_{n - 1}}{{a}_{n}x} + \ldots + \frac{{a}_{2}}{{a}_{n}{x}^{n - 2}} + \frac{{a}_{1}}{{a}_{n}{x}^{n - 1}} + \frac{{a}_{0}}{{a}_{n}{x}^{n}}}\right) \]\n\nAs \( x \rightarrow \pm \infty \), any term with an \( x \) in the denominator becomes closer and closer to 0, and we have \( f\left( x\right) \approx {a}_{n}{x}^{n} \) .
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Yes
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Sketch the graph of \( f\left( x\right) = - 3\left( {{2x} - 1}\right) {\left( x + 1\right) }^{2} \) using end behavior and the multiplicity of its zeros.
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Solution. The end behavior of the graph of \( f \) will match that of its leading term. To find the leading term, we multiply by the leading terms of each factor to get \( \left( {-3}\right) \left( {2x}\right) {\left( x\right) }^{2} = - 6{x}^{3} \) . This tells us that the graph will start above the \( x \) -axis, in Quadrant II, and finish below the \( x \) -axis, in Quadrant IV. Next, we find the zeros of \( f \) . Fortunately for us, \( f \) is factored. \( {}^{15} \) Setting each factor equal to zero gives is \( x = \frac{1}{2} \) and \( x = - 1 \) as zeros. To find the multiplicity of \( x = \frac{1}{2} \) we note that it corresponds to the factor \( \left( {{2x} - 1}\right) \) . This isn’t strictly in the form required in Definition 3.3. If we factor out the 2, however, we get \( \left( {{2x} - 1}\right) = 2\left( {x - \frac{1}{2}}\right) \), and we see that the multiplicity of \( x = \frac{1}{2} \) is 1 . Since 1 is an odd number, we know from Theorem 3.3 that the graph of \( f \) will cross through the \( x \) -axis at \( \left( {\frac{1}{2},0}\right) \) . Since the zero \( x = - 1 \) corresponds to the factor \( {\left( x + 1\right) }^{2} = {\left( x - \left( -1\right) \right) }^{2} \) , we find its multiplicity to be 2 which is an even number. As such, the graph of \( f \) will touch and rebound from the \( x \) -axis at \( \left( {-1,0}\right) \) . Though we’re not asked to, we can find the \( y \) -intercept by finding \( f\left( 0\right) = - 3\left( {2\left( 0\right) - 1}\right) {\left( 0 + 1\right) }^{2} = 3 \) . Thus \( \left( {0,3}\right) \) is an additional point on the graph. Putting this together gives us the graph below.\n\n
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Yes
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Theorem 3.5. The Remainder Theorem: Suppose \( p \) is a polynomial of degree at least 1 and \( c \) is a real number. When \( p\left( x\right) \) is divided by \( x - c \) the remainder is \( p\left( c\right) \) .
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The proof of Theorem 3.5 is a direct consequence of Theorem 3.4. When a polynomial is divided by \( x - c \), the remainder is either 0 or has degree less than the degree of \( x - c \) . Since \( x - c \) is degree 1 , the degree of the remainder must be 0 , which means the remainder is a constant. Hence, in either case, \( p\left( x\right) = \left( {x - c}\right) q\left( x\right) + r \), where \( r \), the remainder, is a real number, possibly 0 . It follows that \( p\left( c\right) = \left( {c - c}\right) q\left( c\right) + r = 0 \cdot q\left( c\right) + r = r \), so we get \( r = p\left( c\right) \) as required.
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Yes
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Theorem 3.6. The Factor Theorem: Suppose \( p \) is a nonzero polynomial. The real number \( c \) is a zero of \( p \) if and only if \( \left( {x - c}\right) \) is a factor of \( p\left( x\right) \) .
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The proof of The Factor Theorem is a consequence of what we already know. If \( \left( {x - c}\right) \) is a factor of \( p\left( x\right) \), this means \( p\left( x\right) = \left( {x - c}\right) q\left( x\right) \) for some polynomial \( q \) . Hence, \( p\left( c\right) = \left( {c - c}\right) q\left( c\right) = 0 \), so \( c \) is a zero of \( p \) . Conversely, if \( c \) is a zero of \( p \), then \( p\left( c\right) = 0 \) . In this case, The Remainder Theorem tells us the remainder when \( p\left( x\right) \) is divided by \( \left( {x - c}\right) \), namely \( p\left( c\right) \), is 0, which means \( \left( {x - c}\right) \) is a factor of \( p \) . What we have established is the fundamental connection between zeros of polynomials and factors of polynomials.
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Yes
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Use synthetic division to perform the following polynomial divisions. Find the quotient and the remainder polynomials, then write the dividend, quotient and remainder in the form given in Theorem 3.4.
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1. When setting up the synthetic division tableau, we need to enter 0 for the coefficient of \( x \) in the dividend. Doing so gives\n\n\n\nSince the dividend was a third degree polynomial, the quotient is a quadratic polynomial with coefficients 5,13 and 39. Our quotient is \( q\left( x\right) = 5{x}^{2} + {13x} + {39} \) and the remainder is \( r\left( x\right) = {118} \) . According to Theorem 3.4, we have \( 5{x}^{3} - 2{x}^{2} + 1 = \left( {x - 3}\right) \left( {5{x}^{2} + {13x} + {39}}\right) + {118} \) .
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Yes
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1. Find \( p\left( {-2}\right) \) using The Remainder Theorem. Check your answer by substitution.
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1. The Remainder Theorem states \( p\left( {-2}\right) \) is the remainder when \( p\left( x\right) \) is divided by \( x - \left( {-2}\right) \) . We set up our synthetic division tableau below. We are careful to record the coefficient of \( {x}^{2} \) as 0 , and proceed as above.\n\n\n\nAccording to the Remainder Theorem, \( p\left( {-2}\right) = - 3 \) . We can check this by direct substitution into the formula for \( p\left( x\right) : p\left( {-2}\right) = 2{\left( -2\right) }^{3} - 5\left( {-2}\right) + 3 = - {16} + {10} + 3 = - 3 \) .
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Yes
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Let \( p\left( x\right) = 4{x}^{4} - 4{x}^{3} - {11}{x}^{2} + {12x} - 3 \) . Given that \( x = \frac{1}{2} \) is a zero of multiplicity 2, find all of the real zeros of \( p \) .
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Solution. We set up for synthetic division. Since we are told the multiplicity of \( \frac{1}{2} \) is two, we continue our tableau and divide \( \frac{1}{2} \) into the quotient polynomial\n\n\n\nFrom the first division, we get \( 4{x}^{4} - 4{x}^{3} - {11}{x}^{2} + {12x} - 3 = \left( {x - \frac{1}{2}}\right) \left( {4{x}^{3} - 2{x}^{2} - {12x} + 6}\right) \) . The second division tells us \( 4{x}^{3} - 2{x}^{2} - {12x} + 6 = \left( {x - \frac{1}{2}}\right) \left( {4{x}^{2} - {12}}\right) \) . Combining these results, we have \( 4{x}^{4} - 4{x}^{3} - {11}{x}^{2} + {12x} - 3 = {\left( x - \frac{1}{2}\right) }^{2}\left( {4{x}^{2} - {12}}\right) \) . To find the remaining zeros of \( p \), we set \( 4{x}^{2} - {12} = 0 \) and get \( x = \pm \sqrt{3} \).
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Yes
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Theorem 3.7. Suppose \( f \) is a polynomial of degree \( n \geq 1 \) . Then \( f \) has at most \( n \) real zeros, counting multiplicities.
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Theorem 3.7 is a consequence of the Factor Theorem and polynomial multiplication. Every zero \( c \) of \( f \) gives us a factor of the form \( \left( {x - c}\right) \) for \( f\left( x\right) \) . Since \( f \) has degree \( n \), there can be at most \( n \) of these factors.
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Yes
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Theorem 3.8. Cauchy’s Bound: Suppose \( f\left( x\right) = {a}_{n}{x}^{n} + {a}_{n - 1}{x}^{n - 1} + \ldots + {a}_{1}x + {a}_{0} \) is a polynomial of degree \( n \) with \( n \geq 1 \) . Let \( M \) be the largest of the numbers: \( \frac{\left| {a}_{0}\right| }{\left| {a}_{n}\right| },\frac{\left| {a}_{1}\right| }{\left| {a}_{n}\right| },\ldots ,\frac{\left| {a}_{n - 1}\right| }{\left| {a}_{n}\right| } \) . Then all the real zeros of \( f \) lie in in the interval \( \left\lbrack {-\left( {M + 1}\right), M + 1}\right\rbrack \) .
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The proof of this fact is not easily explained within the confines of this text. This paper contains the result and gives references to its proof.
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No
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Example 3.3.1. Let \( f\left( x\right) = 2{x}^{4} + 4{x}^{3} - {x}^{2} - {6x} - 3 \) . Determine an interval which contains all of the real zeros of \( f \) .
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Solution. To find the \( M \) stated in Cauchy’s Bound, we take the absolute value of the leading coefficient, in this case \( \left| 2\right| = 2 \) and divide it into the largest (in absolute value) of the remaining coefficients, in this case \( \left| {-6}\right| = 6 \) . This yields \( M = 3 \) so it is guaranteed that all of the real zeros of \( f \) lie in the interval \( \left\lbrack {-4,4}\right\rbrack \) .
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Yes
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Theorem 3.9. Rational Zeros Theorem: Suppose \( f\left( x\right) = {a}_{n}{x}^{n} + {a}_{n - 1}{x}^{n - 1} + \ldots + {a}_{1}x + {a}_{0} \) is a polynomial of degree \( n \) with \( n \geq 1 \), and \( {a}_{0},{a}_{1},\ldots {a}_{n} \) are integers. If \( r \) is a rational zero of \( f \), then \( r \) is of the form \( \pm \frac{p}{q} \), where \( p \) is a factor of the constant term \( {a}_{0} \), and \( q \) is a factor of the leading coefficient \( {a}_{n} \) .
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The Rational Zeros Theorem gives us a list of numbers to try in our synthetic division and that is a lot nicer than simply guessing. If none of the numbers in the list are zeros, then either the polynomial has no real zeros at all, or all of the real zeros are irrational numbers. To see why the Rational Zeros Theorem works, suppose \( c \) is a zero of \( f \) and \( c = \frac{p}{q} \) in lowest terms. This means \( p \) and \( q \) have no common factors. Since \( f\left( c\right) = 0 \), we have\n\n\[ {a}_{n}{\left( \frac{p}{q}\right) }^{n} + {a}_{n - 1}{\left( \frac{p}{q}\right) }^{n - 1} + \ldots + {a}_{1}\left( \frac{p}{q}\right) + {a}_{0} = 0. \]\n\nMultiplying both sides of this equation by \( {q}^{n} \), we clear the denominators to get\n\n\[ {a}_{n}{p}^{n} + {a}_{n - 1}{p}^{n - 1}q + \ldots + {a}_{1}p{q}^{n - 1} + {a}_{0}{q}^{n} = 0 \]\n\nRearranging this equation, we get\n\n\[ {a}_{n}{p}^{n} = - {a}_{n - 1}{p}^{n - 1}q - \ldots - {a}_{1}p{q}^{n - 1} - {a}_{0}{q}^{n} \]\n\nNow, the left hand side is an integer multiple of \( p \), and the right hand side is an integer multiple of \( q \) . (Can you see why?) This means \( {a}_{n}{p}^{n} \) is both a multiple of \( p \) and a multiple of \( q \) . Since \( p \) and \( q \) have no common factors, \( {a}_{n} \) must be a multiple of \( q \) . If we rearrange the equation\n\n\[ {a}_{n}{p}^{n} + {a}_{n - 1}{p}^{n - 1}q + \ldots + {a}_{1}p{q}^{n - 1} + {a}_{0}{q}^{n} = 0 \]\n\nas\n\n\[ {a}_{0}{q}^{n} = - {a}_{n}{p}^{n} - {a}_{n - 1}{p}^{n - 1}q - \ldots - {a}_{1}p{q}^{n - 1} \]\n\nwe can play the same game and conclude \( {a}_{0} \) is a multiple of \( p \), and we have the result.
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Yes
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Let \( f\left( x\right) = 2{x}^{4} + 4{x}^{3} - {x}^{2} - {6x} - 3 \) . Use the Rational Zeros Theorem to list all of the possible rational zeros of \( f \) .
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Solution. To generate a complete list of rational zeros, we need to take each of the factors of constant term, \( {a}_{0} = - 3 \), and divide them by each of the factors of the leading coefficient \( {a}_{4} = 2 \) . The factors of \( - 3 \) are \( \pm 1 \) and \( \pm 3 \) . Since the Rational Zeros Theorem tacks on a \( \pm \) anyway, for the moment, we consider only the positive factors 1 and 3 . The factors of 2 are 1 and 2, so the Rational Zeros Theorem gives the list \( \left\{ {\pm \frac{1}{1}, \pm \frac{1}{2}, \pm \frac{3}{1}, \pm \frac{3}{2}}\right\} \) or \( \left\{ {\pm \frac{1}{2}, \pm 1, \pm \frac{3}{2}, \pm 3}\right\} \) .
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Yes
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Let \( f\left( x\right) = 2{x}^{4} + 4{x}^{3} - {x}^{2} - {6x} - 3 \) .\n\n1. Graph \( y = f\left( x\right) \) on the calculator using the interval obtained in Example 3.3.1 as a guide.\n\n2. Use the graph to shorten the list of possible rational zeros obtained in Example 3.3.2.\n\n3. Use synthetic division to find the real zeros of \( f \), and state their multiplicities.
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## Solution.\n\n1. In Example 3.3.1, we determined all of the real zeros of \( f \) lie in the interval \( \left\lbrack {-4,4}\right\rbrack \) . We set our window accordingly and get\n\n\n\n2. In Example 3.3.2, we learned that any rational zero of \( f \) must be in the list \( \left\{ {\pm \frac{1}{2}, \pm 1, \pm \frac{3}{2}, \pm 3}\right\} \) . From the graph, it looks as if we can rule out any of the positive rational zeros, since the graph seems to cross the \( x \) -axis at a value just a little greater than 1 . On the negative side, -1 looks good, so we try that for our synthetic division.\n\n\n\nWe have a winner! Remembering that \( f \) was a fourth degree polynomial, we know that our quotient is a third degree polynomial. If we can do one more successful division, we will have knocked the quotient down to a quadratic, and, if all else fails, we can use the quadratic formula to find the last two zeros. Since there seems to be no other rational zeros to try, we continue with -1 . Also, the shape of the crossing at \( x = - 1 \) leads us to wonder if the zero \( x = - 1 \) has multiplicity 3 .\n\n\n\nSuccess! Our quotient polynomial is now \( 2{x}^{2} - 3 \) . Setting this to zero gives \( 2{x}^{2} - 3 = 0 \), or \( {x}^{2} = \frac{3}{2} \), which gives us \( x = \pm \frac{\sqrt{6}}{2} \) . Concerning multiplicities, based on our division, we have that -1 has a multiplicity of at least 2 . The Factor Theorem tells us our remaining zeros, \( \pm \frac{\sqrt{6}}{2} \), each have multiplicity at least 1 . However, Theorem 3.7 tells us \( f \) can have at most 4 real zeros, counting multiplicity, and so we conclude that -1 is of multiplicity exactly 2 and \( \pm \frac{\sqrt{6}}{2} \) each has multiplicity 1. (Thus, we were wrong to think that -1 had multiplicity 3.)
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Yes
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Use Cauchy’s Bound to determine an interval in which all of the real zeros of \( f \) lie.
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Applying Cauchy’s Bound, we find \( M = {12} \), so all of the real zeros lie in the interval \( \left\lbrack {-{13},{13}}\right\rbrack \) .
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Yes
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Let \( f\left( x\right) = 2{x}^{4} + 4{x}^{3} - {x}^{2} - {6x} - 3 \) . Use Descartes’ Rule of Signs to determine the possible number and location of the real zeros of \( f \) .
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Solution. As noted above, the variations of sign of \( f\left( x\right) \) is 1 . This means, counting multiplicities, \( f \) has exactly 1 positive real zero. Since \( f\left( {-x}\right) = 2{\left( -x\right) }^{4} + 4{\left( -x\right) }^{3} - {\left( -x\right) }^{2} - 6\left( {-x}\right) - 3 = \) \( 2{x}^{4} - 4{x}^{3} - {x}^{2} + {6x} - 3 \) has 3 variations in sign, \( f \) has either 3 negative real zeros or 1 negative real zero, counting multiplicities.
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Yes
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Theorem 3.11. Upper and Lower Bounds: Suppose \( f \) is a polynomial of degree \( n \geq 1 \) . - If \( c > 0 \) is synthetically divided into \( f \) and all of the numbers in the final line of the division tableau have the same signs, then \( c \) is an upper bound for the real zeros of \( f \) . That is, there are no real zeros greater than \( c \) . - If \( c < 0 \) is synthetically divided into \( f \) and the numbers in the final line of the division tableau alternate signs, then \( c \) is a lower bound for the real zeros of \( f \) . That is, there are no real zeros less than \( c \) . NOTE: If the number 0 occurs in the final line of the division tableau in either of the above cases, it can be treated as \( \left( +\right) \) or \( \left( -\right) \) as needed.
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The Upper and Lower Bounds Theorem works because of Theorem 3.4. For the upper bound part of the theorem, suppose \( c > 0 \) is divided into \( f \) and the resulting line in the division tableau contains, for example, all nonnegative numbers. This means \( f\left( x\right) = \left( {x - c}\right) q\left( x\right) + r \), where the coefficients of the quotient polynomial and the remainder are nonnegative. (Note that the leading coefficient of \( q \) is the same as \( f \) so \( q\left( x\right) \) is not the zero polynomial.) If \( b > c \), then \( f\left( b\right) = \left( {b - c}\right) q\left( b\right) + r \) , where \( \left( {b - c}\right) \) and \( q\left( b\right) \) are both positive and \( r \geq 0 \) . Hence \( f\left( b\right) > 0 \) which shows \( b \) cannot be a zero of \( f \) . Thus no real number \( b > c \) can be a zero of \( f \), as required. A similar argument proves \( f\left( b\right) < 0 \) if all of the numbers in the final line of the synthetic division tableau are non-positive. To prove the lower bound part of the theorem, we note that a lower bound for the negative real zeros of \( f\left( x\right) \) is an upper bound for the positive real zeros of \( f\left( {-x}\right) \) . Applying the upper bound portion to \( f\left( {-x}\right) \) gives the result. (Do you see where the alternating signs come in?)
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Yes
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1. Find all of the real zeros of \( f \) and their multiplicities.
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We know from Cauchy’s Bound that all of the real zeros lie in the interval \( \left\lbrack {-4,4}\right\rbrack \) and that our possible rational zeros are \( \pm \frac{1}{2}, \pm 1, \pm \frac{3}{2} \) and \( \pm 3 \) . Descartes’ Rule of Signs guarantees us at least one negative real zero and exactly one positive real zero, counting multiplicity. We try our positive rational zeros, starting with the smallest, \( \frac{1}{2} \) . Since the remainder isn’t zero, we know \( \frac{1}{2} \) isn’t a zero. Sadly, the final line in the division tableau has both positive and negative numbers, so \( \frac{1}{2} \) is not an upper bound. The only information we get from this division is courtesy of the Remainder Theorem which tells us \( f\left( \frac{1}{2}\right) = - \frac{45}{8} \) so the point \( \left( {\frac{1}{2}, - \frac{45}{8}}\right) \) is on the graph of \( f \) . We continue to our next possible zero,1 . As before, the only information we can glean from this is that \( \left( {1, - 4}\right) \) is on the graph of \( f \) . When we try our next possible zero, \( \frac{3}{2} \), we get that it is not a zero, and we also see that it is an upper bound on the zeros of \( f \), since all of the numbers in the final line of the division tableau are positive. This means there is no point trying our last possible rational zero, 3. Descartes' Rule of Signs guaranteed us a positive real zero, and at this point we have shown this zero is irrational. Furthermore, the Intermediate Value Theorem, Theorem 3.1, tells us the zero lies between 1 and \( \frac{3}{2} \), since \( f\left( 1\right) < 0 \) and \( f\left( \frac{3}{2}\right) > 0 \) . We now turn our attention to negative real zeros. We try the largest possible zero, \( - \frac{1}{2} \) . Synthetic division shows us it is not a zero, nor is it a lower bound (since the numbers in the final line of the division tableau do not alternate), so we proceed to -1 . This division shows -1 is a zero. Descartes' Rule of Signs told us that we may have up to three negative real zeros, counting multiplicity, so we try -1 again, and it works once more. At this point, we have taken \( f \), a fourth degree polynomial, and performed two successful divisions. Our quotient polynomial is quadratic, so we look at it to find the remaining zeros. Setting the quotient polynomial equal to zero yields \( 2{x}^{2} - 3 = 0 \), so that \( {x}^{2} = \frac{3}{2} \), or \( x = \pm \frac{\sqrt{6}}{2} \) . Descartes’ Rule of Signs tells us that the positive real zero we found, \( \frac{\sqrt{6}}{2} \), has multiplicity 1 . Descartes also tells us the total multiplicity of negative real zeros is 3 , which forces -1 to be a zero of multiplicity 2 and \( - \frac{\sqrt{6}}{2} \) to have multiplicity 1 .
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Yes
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Example 3.3.8. Suppose the profit \( P \), in thousands of dollars, from producing and selling \( x \) hundred LCD TVs is given by \( P\left( x\right) = - 5{x}^{3} + {35}{x}^{2} - {45x} - {25},0 \leq x \leq {10.07} \) . How many TVs should be produced to make a profit? Check your answer using a graphing utility.
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Solution. To ’make a profit’ means to solve \( P\left( x\right) = - 5{x}^{3} + {35}{x}^{2} - {45x} - {25} > 0 \), which we do analytically using a sign diagram. To simplify things, we first factor out the -5 common to all the coefficients to get \( - 5\left( {{x}^{3} - 7{x}^{2} + {9x} - 5}\right) > 0 \), so we can just focus on finding the zeros of \( f\left( x\right) = {x}^{3} - 7{x}^{2} + {9x} + 5 \) . The possible rational zeros of \( f \) are \( \pm 1 \) and \( \pm 5 \), and going through the usual computations, we find \( x = 5 \) is the only rational zero. Using this, we factor \( f\left( x\right) = {x}^{3} - 7{x}^{2} + {9x} + 5 = \left( {x - 5}\right) \left( {{x}^{2} - {2x} - 1}\right) \), and we find the remaining zeros by applying the Quadratic Formula to \( {x}^{2} - {2x} - 1 = 0 \) . We find three real zeros, \( x = 1 - \sqrt{2} = - {0.414}\ldots \) , \( x = 1 + \sqrt{2} = {2.414}\ldots \), and \( x = 5 \), of which only the last two fall in the applied domain of \( \left\lbrack {0,{10.07}}\right\rbrack \) . We choose \( x = 0, x = 3 \) and \( x = {10.07} \) as our test values and plug them into the function \( P\left( x\right) = - 5{x}^{3} + {35}{x}^{2} - {45x} - {25} \) (not \( f\left( x\right) = {x}^{3} - 7{x}^{2} + {9x} - 5 \) ) to get the sign diagram below.\n\n\n\nWe see immediately that \( P\left( x\right) > 0 \) on \( \left( {1 + \sqrt{2},5}\right) \) . Since \( x \) measures the number of TVs in hundreds, \( x = 1 + \sqrt{2} \) corresponds to \( {241.4}\ldots \) TVs. Since we can’t produce a fractional part of a TV, we need to choose between producing 241 and 242 TVs. From the sign diagram, we see that \( P\left( {2.41}\right) < 0 \) but \( P\left( {2.42}\right) > 0 \) so, in this case we take the next larger integer value and set the minimum production to \( {242}\mathrm{{TVs}} \) . At the other end of the interval, we have \( x = 5 \) which corresponds to \( {500}\mathrm{{TVs}} \) . Here, we take the next smaller integer value, 499 TVs to ensure that we make a profit. Hence, in order to make a profit, at least 242, but no more than 499 TVs need to be produced. To check our answer using a calculator, we graph \( y = P\left( x\right) \) and make use of the ’Zero’ command. We see that the calculator approximations bear out our analysis. \( {}^{7} \)
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Yes
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Perform the indicated operations. Write your answer in the form \( {}^{5}a + {bi} \) .
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1. As mentioned earlier, we treat expressions involving \( i \) as we would any other radical. We combine like terms to get \( \left( {1 - {2i}}\right) - \left( {3 + {4i}}\right) = 1 - {2i} - 3 - {4i} = - 2 - {6i} \) .
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No
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Theorem 3.12. Properties of the Complex Conjugate: Let \( z \) and \( w \) be complex numbers.\n\n- \( \overline{\bar{z}} = z \)\n\n- \( \bar{z} + \bar{w} = \overline{z + w} \)\n\n- \( \bar{z}\bar{w} = \overline{zw} \)\n\n- \( {\left( \bar{z}\right) }^{n} = \overline{{z}^{n}} \), for any natural number \( n \)\n\n- \( z \) is a real number if and only if \( \bar{z} = z \) .
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Essentially, Theorem 3.12 says that complex conjugation works well with addition, multiplication and powers. The proof of these properties can best be achieved by writing out \( z = a + {bi} \) and \( w = c + {di} \) for real numbers \( a, b, c \) and \( d \) . Next, we compute the left and right hand sides of each equation and check to see that they are the same. The proof of the first property is a very quick exercise. \( {}^{7} \) To prove the second property, we compare \( \bar{z} + \bar{w} \) and \( \overline{z + w} \) . We have \( \bar{z} + \bar{w} = \overline{a + {bi}} + \overline{c + {di}} = a - {bi} + c - {di} \) . To find \( \overline{z + w} \), we first compute\n\n\[ z + w = \left( {a + {bi}}\right) + \left( {c + {di}}\right) = \left( {a + c}\right) + \left( {b + d}\right) i \]\n\nso\n\n\[ \overline{z + w} = \overline{\left( {a + c}\right) + \left( {b + d}\right) i} = \left( {a + c}\right) - \left( {b + d}\right) i = a - {bi} + c - {di} \]\n\nAs such, we have established \( \bar{z} + \bar{w} = \overline{z + w} \) . The proof for multiplication works similarly. The proof that the conjugate works well with powers can be viewed as a repeated application of the product rule, and is best proved using a technique called Mathematical Induction. \( {}^{8} \) The last property is a characterization of real numbers. If \( z \) is real, then \( z = a + {0i} \), so \( \bar{z} = a - {0i} = a = z \) . On the other hand, if \( z = \bar{z} \), then \( a + {bi} = a - {bi} \) which means \( b = - b \) so \( b = 0 \) . Hence, \( z = a + {0i} = a \) and is real.
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No
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1. Find all of the complex zeros of \( f \) and state their multiplicities.
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Since \( f \) is a fifth degree polynomial, we know that we need to perform at least three successful divisions to get the quotient down to a quadratic function. At that point, we can find the remaining zeros using the Quadratic Formula, if necessary. Using the techniques developed in Section 3.3, we get\n\n\n\nOur quotient is \( {12}{x}^{2} - {12x} + {12} \), whose zeros we find to be \( \frac{1 \pm i\sqrt{3}}{2} \) . From Theorem 3.14, we know \( f \) has exactly 5 zeros, counting multiplicities, and as such we have the zero \( \frac{1}{2} \) with multiplicity 2, and the zeros \( - \frac{1}{3},\frac{1 + i\sqrt{3}}{2} \) and \( \frac{1 - i\sqrt{3}}{2} \), each of multiplicity 1 .
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Yes
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Theorem 3.15. Conjugate Pairs Theorem: If \( f \) is a polynomial function with real number coefficients and \( z \) is a zero of \( f \), then so is \( \bar{z} \) .
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To prove the theorem, suppose \( f \) is a polynomial with real number coefficients. Specifically, let \( f\left( x\right) = {a}_{n}{x}^{n} + {a}_{n - 1}{x}^{n - 1} + \ldots + {a}_{2}{x}^{2} + {a}_{1}x + {a}_{0} \) . If \( z \) is a zero of \( f \), then \( f\left( z\right) = 0 \), which means \( {a}_{n}{z}^{n} + {a}_{n - 1}{z}^{n - 1} + \ldots + {a}_{2}{z}^{2} + {a}_{1}z + {a}_{0} = 0 \) . Next, we consider \( f\left( \bar{z}\right) \) and apply Theorem 3.12 below.\n\n\[ f\left( \bar{z}\right) = {a}_{n}{\left( \bar{z}\right) }^{n} + {a}_{n - 1}{\left( \bar{z}\right) }^{n - 1} + \ldots + {a}_{2}{\left( \bar{z}\right) }^{2} + {a}_{1}\bar{z} + {a}_{0} \]\n\n\[ = {a}_{n}\overline{{z}^{n}} + {a}_{n - 1}\overline{{z}^{n - 1}} + \ldots + {a}_{2}\overline{{z}^{2}} + {a}_{1}\bar{z} + {a}_{0} \]\nsince \( {\left( \bar{z}\right) }^{n} = \overline{{z}^{n}} \)\n\n\[ = \overline{{a}_{n}}\overline{{z}^{n}} + \overline{{a}_{n - 1}}\overline{{z}^{n - 1}} + \ldots + \overline{{a}_{2}}\overline{{z}^{2}} + \overline{{a}_{1}}\overline{z} + \overline{{a}_{0}} \]\nsince the coefficients are real\n\n\[ = \overline{{a}_{n}{z}^{n}} + \overline{{a}_{n - 1}{z}^{n - 1}} + \ldots + \overline{{a}_{2}{z}^{2}} + \overline{{a}_{1}z} + \overline{{a}_{0}} \]\nsince \( \bar{z}\bar{w} = \overline{zw} \)\n\n\[ = \overline{{a}_{n}{z}^{n} + {a}_{n - 1}{z}^{n - 1} + \ldots + {a}_{2}{z}^{2} + {a}_{1}z + {a}_{0}} \]\nsince \( \bar{z} + \bar{w} = \overline{z + w} \)\n\n\[ = \overline{f\left( z\right) } \]\n\n\[ = \overline{0} \]\n\n\[ = 0 \]\n\nThis shows that \( \bar{z} \) is a zero of \( f \) . So, if \( f \) is a polynomial function with real number coefficients, Theorem 3.15 tells us that if \( a + {bi} \) is a nonreal zero of \( f \), then so is \( a - {bi} \) . In other words, nonreal zeros of \( f \) come in conjugate pairs. The Factor Theorem kicks in to give us both \( \left( {x - \left\lbrack {a + {bi}}\right\rbrack }\right) \) and \( \left( {x - \left\lbrack {a - {bi}}\right\rbrack }\right) \) as factors of \( f\left( x\right) \) which means \( \left( {x - \left\lbrack {a + {bi}}\right\rbrack }\right) \left( {x - \left\lbrack {a - {bi}}\right\rbrack }\right) = {x}^{2} + {2ax} + \left( {{a}^{2} + {b}^{2}}\right) \) is an irreducible quadratic factor of \( f \) . As a result, we have our last theorem of the section.
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Yes
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Let \( f\left( x\right) = {x}^{4} + {64} \). Use synthetic division to show that \( x = 2 + {2i} \) is a zero of \( f \).
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Remembering to insert the 0 's in the synthetic division tableau we have\n\n
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No
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Find a polynomial \( p \) of lowest degree that has integer coefficients and satisfies all of the following criteria:\n\n- the graph of \( y = p\\left( x\\right) \) touches (but doesn’t cross) the \( x \) -axis at \( \\left( {\\frac{1}{3},0}\\right) \)\n\n- \( x = {3i} \) is a zero of \( p \).\n\n- as \( x \\rightarrow - \\infty, p\\left( x\\right) \\rightarrow - \\infty \)\n\n- as \( x \\rightarrow \\infty, p\\left( x\\right) \\rightarrow - \\infty \)
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Solution. To solve this problem, we will need a good understanding of the relationship between the \( x \) -intercepts of the graph of a function and the zeros of a function, the Factor Theorem, the role of multiplicity, complex conjugates, the Complex Factorization Theorem, and end behavior of polynomial functions. (In short, you'll need most of the major concepts of this chapter.) Since the graph of \( p \) touches the \( x \) -axis at \( \\left( {\\frac{1}{3},0}\\right) \), we know \( x = \\frac{1}{3} \) is a zero of even multiplicity. Since we are after a polynomial of lowest degree, we need \( x = \\frac{1}{3} \) to have multiplicity exactly 2 . The Factor Theorem now tells us \( {\\left( x - \\frac{1}{3}\\right) }^{2} \) is a factor of \( p\\left( x\\right) \) . Since \( x = {3i} \) is a zero and our final answer is to have integer (real) coefficients, \( x = - {3i} \) is also a zero. The Factor Theorem kicks in again to give us \( \\left( {x - {3i}}\\right) \) and \( \\left( {x + {3i}}\\right) \) as factors of \( p\\left( x\\right) \) . We are given no further information about zeros or intercepts so we conclude, by the Complex Factorization Theorem that \( p\\left( x\\right) = a{\\left( x - \\frac{1}{3}\\right) }^{2}\\left( {x - {3i}}\\right) \\left( {x + {3i}}\\right) \) for some real number \( a \) . Expanding this, we get \( p\\left( x\\right) = a{x}^{4} - \\frac{2a}{3}{x}^{3} + \\frac{82a}{9}{x}^{2} - {6ax} + a \) . In order to obtain integer coefficients, we know a must be an integer multiple of 9 . Our last concern is end behavior. Since the leading term of \( p\\left( x\\right) \) is \( a{x}^{4} \), we need \( a < 0 \) to get \( p\\left( x\\right) \\rightarrow - \\infty \) as \( x \\rightarrow \\pm \\infty \) . Hence, if we choose \( x = - 9 \), we get \( p\\left( x\\right) = - 9{x}^{4} + 6{x}^{3} - {82}{x}^{2} + {54x} - 9 \) . We can verify our handiwork using the techniques developed in this chapter.
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Yes
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Find the domain of the following rational functions. Write them in the form \( \frac{p\left( x\right) }{q\left( x\right) } \) for polynomial functions \( p \) and \( q \) and simplify.
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1. To find the domain of \( f \), we proceed as we did in Section 1.4: we find the zeros of the denominator and exclude them from the domain. Setting \( x + 1 = 0 \) results in \( x = - 1 \) . Hence, our domain is \( \left( {-\infty , - 1}\right) \cup \left( {-1,\infty }\right) \) . The expression \( f\left( x\right) \) is already in the form requested and when we check for common factors among the numerator and denominator we find none, so we are done.
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No
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Theorem 4.1. Location of Vertical Asymptotes and Holes: \( {}^{a} \) Suppose \( r \) is a rational function which can be written as \( r\left( x\right) = \frac{p\left( x\right) }{q\left( x\right) } \) where \( p \) and \( q \) have no common zeros. \( {}^{b} \) Let \( c \) be a real number which is not in the domain of \( r \) .
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- If \( q\left( c\right) \neq 0 \), then the graph of \( y = r\left( x\right) \) has a hole at \( \left( {c,\frac{p\left( c\right) }{q\left( c\right) }}\right) \).\n\n- If \( q\left( c\right) = 0 \), then the line \( x = c \) is a vertical asymptote of the graph of \( y = r\left( x\right) \) .
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Yes
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Find the vertical asymptotes of, and/or holes in, the graphs of the following rational functions. Verify your answers using a graphing calculator, and describe the behavior of the graph near them using proper notation.
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1. To use Theorem 4.1, we first find all of the real numbers which aren’t in the domain of \( f \) . To do so, we solve \( {x}^{2} - 3 = 0 \) and get \( x = \pm \sqrt{3} \) . Since the expression \( f\left( x\right) \) is in lowest terms, there is no cancellation possible, and we conclude that the lines \( x = - \sqrt{3} \) and \( x = \sqrt{3} \) are vertical asymptotes to the graph of \( y = f\left( x\right) \) . The calculator verifies this claim, and from the graph, we see that as \( x \rightarrow - {\sqrt{3}}^{ - }, f\left( x\right) \rightarrow - \infty \), as \( x \rightarrow - {\sqrt{3}}^{ + }, f\left( x\right) \rightarrow \infty \), as \( x \rightarrow {\sqrt{3}}^{ - } \) , \( f\left( x\right) \rightarrow - \infty \), and finally as \( x \rightarrow {\sqrt{3}}^{ + }, f\left( x\right) \rightarrow \infty \) .
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Yes
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1. Find and interpret \( P\left( 0\right) \) .
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Substituting \( t = 0 \) gives \( P\left( 0\right) = \frac{100}{{\left( 5 - 0\right) }^{2}} = 4 \), which means 4000 bacteria are initially introduced into the environment.
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Yes
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List the horizontal asymptotes, if any, of the graphs of the following functions. Verify your answers using a graphing calculator, and describe the behavior of the graph near them using proper notation.
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1. The numerator of \( f\left( x\right) \) is \( {5x} \), which has degree 1 . The denominator of \( f\left( x\right) \) is \( {x}^{2} + 1 \), which has degree 2. Applying Theorem 4.2, \( y = 0 \) is the horizontal asymptote. Sure enough, we see from the graph that as \( x \rightarrow - \infty, f\left( x\right) \rightarrow {0}^{ - } \) and as \( x \rightarrow \infty, f\left( x\right) \rightarrow {0}^{ + } \) .
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Yes
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Example 4.1.5. The number of students \( N \) at local college who have had the flu \( t \) months after the semester begins can be modeled by the formula \( N\left( t\right) = {500} - \frac{450}{1 + {3t}} \) for \( t \geq 0 \).
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## Solution.\n\n1. \( N\left( 0\right) = {500} - \frac{450}{1 + 3\left( 0\right) } = {50} \) . This means that at the beginning of the semester,50 students\n\nhave had the flu.\n\n2. We set \( N\left( t\right) = {300} \) to get \( {500} - \frac{450}{1 + {3t}} = {300} \) and solve. Isolating the fraction gives \( \frac{450}{1 + {3t}} = {200} \) . Clearing denominators gives \( {450} = {200}\left( {1 + {3t}}\right) \) . Finally, we get \( t = \frac{5}{12} \) . This means it will take \( \frac{5}{12} \) months, or about 13 days, for 300 students to have had the flu.\n\n3. To determine the behavior of \( N \) as \( t \rightarrow \infty \), we can use a table.\n\n<table><thead><tr><th>\( t \)</th><th>\( N\left( t\right) \)</th></tr></thead><tr><td>10</td><td>\( \approx {485.48} \)</td></tr><tr><td>100</td><td>\( \approx {498.50} \)</td></tr><tr><td>1000</td><td>\( \approx {499.85} \)</td></tr><tr><td>10000</td><td>\( \approx {499.98} \)</td></tr></table>\n\nThe table suggests that as \( t \rightarrow \infty, N\left( t\right) \rightarrow {500} \) . (More specifically, \( {500}^{ - } \) .) This means as time goes by, only a total of 500 students will have ever had the flu.
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Yes
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Theorem 4.3. Determination of Slant Asymptotes: Suppose \( r \) is a rational function and \( r\left( x\right) = \frac{p\left( x\right) }{q\left( x\right) } \), where the degree of \( p \) is exactly one more than the degree of \( q \) . Then the graph of \( y = r\left( x\right) \) has the slant asymptote \( y = L\left( x\right) \) where \( L\left( x\right) \) is the quotient obtained by dividing \( p\left( x\right) \) by \( q\left( x\right) \) .
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In the same way that Theorem 4.2 gives us an easy way to see if the graph of a rational function \( r\left( x\right) = \frac{p\left( x\right) }{q\left( x\right) } \) has a horizontal asymptote by comparing the degrees of the numerator and denominator, Theorem 4.3 gives us an easy way to check for slant asymptotes. Unlike Theorem 4.2, which gives us a quick way to find the horizontal asymptotes (if any exist), Theorem 4.3 gives us no such ’short-cut’. If a slant asymptote exists, we have no recourse but to use long division to find it.
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Yes
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Find the slant asymptotes of the graphs of the following functions if they exist. Verify your answers using a graphing calculator and describe the behavior of the graph near them using proper notation.
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1. The degree of the numerator is 2 and the degree of the denominator is 1 , so Theorem 4.3 guarantees us a slant asymptote. To find it, we divide \( 1 - x = - x + 1 \) into \( {x}^{2} - {4x} + 2 \) and get a quotient of \( - x + 3 \), so our slant asymptote is \( y = - x + 3 \) . We confirm this graphically, and we see that as \( x \rightarrow - \infty \), the graph of \( y = f\left( x\right) \) approaches the asymptote from below, and as \( x \rightarrow \infty \), the graph of \( y = f\left( x\right) \) approaches the asymptote from above. \( {}^{16} \)
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Yes
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Sketch the graph of \( r\left( x\right) = \frac{{x}^{4} + 1}{{x}^{2} + 1} \) .
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Solution.\n\n1. The denominator \( {x}^{2} + 1 \) is never zero so the domain is \( \left( {-\infty ,\infty }\right) \).\n\n2. With no real zeros in the denominator, \( {x}^{2} + 1 \) is an irreducible quadratic. Our only hope of reducing \( r\left( x\right) \) is if \( {x}^{2} + 1 \) is a factor of \( {x}^{4} + 1 \) . Performing long division gives us\n\n\[ \frac{{x}^{4} + 1}{{x}^{2} + 1} = {x}^{2} - 1 + \frac{2}{{x}^{2} + 1} \]\n\nThe remainder is not zero so \( r\left( x\right) \) is already reduced.\n\n3. To find the \( x \) -intercept, we’d set \( r\left( x\right) = 0 \) . Since there are no real solutions to \( \frac{{x}^{4} + 1}{{x}^{2} + 1} = 0 \), we have no \( x \) -intercepts. Since \( r\left( 0\right) = 1 \), we get \( \left( {0,1}\right) \) as the \( y \) -intercept.\n\n4. This step doesn’t apply to \( r \), since its domain is all real numbers.\n\n5. For end behavior, we note that since the degree of the numerator is exactly two more than the degree of the denominator, neither Theorems 4.2 nor 4.3 apply. \( {}^{15} \) We know from our attempt to reduce \( r\left( x\right) \) that we can rewrite \( r\left( x\right) = {x}^{2} - 1 + \frac{2}{{x}^{2} + 1} \), so we focus our attention on the term corresponding to the remainder, \( \frac{2}{{x}^{2} + 1} \) It should be clear that as \( x \rightarrow \pm \infty \) , \( \frac{2}{{x}^{2} + 1} \approx \) very small \( \left( +\right) \), which means \( r\left( x\right) \approx {x}^{2} - 1 + \) very small \( \left( +\right) \) . So the graph \( y = r\left( x\right) \) is a little bit above the graph of the parabola \( y = {x}^{2} - 1 \) as \( x \rightarrow \pm \infty \) . Graphically,\n\n\n\n6. There isn’t much work to do for a sign diagram for \( r\left( x\right) \), since its domain is all real numbers and it has no zeros. Our sole test interval is \( \left( {-\infty ,\infty }\right) \), and since we know \( r\left( 0\right) = 1 \), we conclude \( r\left( x\right) \) is \( \left( +\right) \) for all real numbers. At this point, we don’t have much to go on for\n\n---
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No
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1. Solve \( \frac{{x}^{3} - {2x} + 1}{x - 1} = \frac{1}{2}x - 1 \) .
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To solve the equation, we clear denominators\n\n\[ \frac{{x}^{3} - {2x} + 1}{x - 1} = \frac{1}{2}x - 1 \]\n\n\[ \left( \frac{{x}^{3} - {2x} + 1}{x - 1}\right) \cdot 2\left( {x - 1}\right) = \left( {\frac{1}{2}x - 1}\right) \cdot 2\left( {x - 1}\right) \]\n\n\[ 2{x}^{3} - {4x} + 2 = {x}^{2} - {3x} + 2 \]\n\n\[ 2{x}^{3} - {x}^{2} - x = 0 \]\n\n\[ x\left( {{2x} + 1}\right) \left( {x - 1}\right) = 0 \]\n\n\[ x = - \frac{1}{2},0,1 \]\n\nSince we cleared denominators, we need to check for extraneous solutions. Sure enough, we see that \( x = 1 \) does not satisfy the original equation and must be discarded. Our solutions are \( x = - \frac{1}{2} \) and \( x = 0 \) .
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Yes
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Working alone, Taylor can weed the garden in 4 hours. If Carl helps, they can weed the garden in 3 hours. How long would it take for Carl to weed the garden on his own?
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Solution. The key relationship between work and time which we use in this problem is:\namount of work done \( = \) rate of work \( \cdot \) time spent working\n\nWe are told that, working alone, Taylor can weed the garden in 4 hours. In Taylor's case then:\namount of work Taylor does \( = \) rate of Taylor working \( \cdot \) time Taylor spent working\n1 garden \( = \) (rate of Taylor working) \( \cdot \left( {4\text{ hours}}\right) \)\nSo we have that the rate Taylor works is \( \frac{1\text{ garden }}{4\text{ hours }} = \frac{1}{4}\frac{\text{ garden }}{\text{ hour }} \). We are also told that when working together, Taylor and Carl can weed the garden in just 3 hours. We have:\namount of work done together \( = \) rate of working together \( \cdot \) time spent working together\n1 garden \( = \) (rate of working together) \( \cdot \) (3 hours)\nFrom this, we find that the rate of Taylor and Carl working together is \( \frac{1\text{ garden }}{3\text{ hours }} = \frac{1}{3}\frac{\text{garden }}{\text{ hour }} \). We are asked to find out how long it would take for Carl to weed the garden on his own. Let us call this unknown \( t \), measured in hours to be consistent with the other times given to us in the problem. Then:\namount of work Carl does \( = \) rate of Carl working \( \cdot \) time Carl spent working\n\[ \text{1 garden} = \text{(rate of Carl working)} \cdot \text{(thours)} \]\nIn order to find \( t \), we need to find the rate of Carl working, so let’s call this quantity \( R \), with units \( \frac{\text{ garden }}{\text{ hour }} \). Using the fact that rates are additive, we have:\n\[ \text{rate working together} = \text{rate of Taylor working + rate of Carl working} \]\n\[ \frac{1}{3}\frac{\text{ garden }}{\text{ hour }} = \frac{1}{4}\frac{\text{ garden }}{\text{ hour }} + R\frac{\text{ garden }}{\text{ hour }} \]\nso that \( R = \frac{1}{12}\frac{\text{garden}}{\text{hour}} \). Substituting this into our ’work-rate-time’ equation for Carl, we get:\n\[ \text{1 garden} = \text{(rate of Carl working)} \cdot \text{(}t\text{hours)} \]\n\[ \text{1 garden} = \left( {\frac{1}{12}\frac{\text{ garden }}{\text{ hour }}}\right) \cdot \left( {t\text{ hours }}\right) \]\nSolving \( 1 = \frac{1}{12}t \), we get \( t = {12} \), so it takes Carl 12 hours to weed the garden on his own.
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Yes
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1. Find an expression for the average cost function \( \bar{C}\left( x\right) \) .
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1. From \( \bar{C}\left( x\right) = \frac{C\left( x\right) }{x} \), we obtain \( \bar{C}\left( x\right) = \frac{{80x} + {150}}{x} \). The domain of \( C \) is \( x \geq 0 \), but since \( x = 0 \) causes problems for \( \bar{C}\left( x\right) \), we get our domain to be \( x > 0 \), or \( \left( {0,\infty }\right) \) .
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Yes
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1. Express the height \( h \) in centimeters as a function of the width \( x \) and state the applied domain.
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1. We are told that the volume of the box is 1000 cubic centimeters and that \( x \) represents the width, in centimeters. From geometry, we know Volume \( = \) width \( \times \) height \( \times \) depth. Since the base of the box is a square, the width and the depth are both \( x \) centimeters. Using \( h \) for the height, we have \( {1000} = {x}^{2}h \), so that \( h = \frac{1000}{{x}^{2}} \) . Using function notation, \( {}^{8}h\left( x\right) = \frac{1000}{{x}^{2}} \) As for the applied domain, in order for there to be a box at all, \( x > 0 \), and since every such choice of \( x \) will return a positive number for the height \( h \) we have no other restrictions and conclude our domain is \( \left( {0,\infty }\right) \) .
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Yes
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1. Hooke’s Law: The force \( F \) exerted on a spring is directly proportional the extension \( x \) of the spring.
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1. Applying the definition of direct variation, we get \( F = {kx} \) for some constant \( k \) .
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Yes
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1. Use your calculator to generate a scatter diagram for these data using \( V \) as the independent variable and \( P \) as the dependent variable. Does it appear from the graph that \( P \) is inversely proportional to \( V \) ? Explain.
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1. If \( P \) really does vary inversely with \( V \), then \( P = \frac{k}{V} \) for some constant \( k \) . From the data plot, the points do seem to lie along a curve like \( y = \frac{k}{x} \) .
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No
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1. \( \left( {g \circ f}\right) \left( 1\right) \)
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Using Definition 5.1, \( \left( {g \circ f}\right) \left( 1\right) = g\left( {f\left( 1\right) }\right) \) . We find \( f\left( 1\right) = - 3 \), so\n\n\[ \left( {g \circ f}\right) \left( 1\right) = g\left( {f\left( 1\right) }\right) = g\left( {-3}\right) = 2 \]
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Yes
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Theorem 5.1. Properties of Function Composition: Suppose \( f, g \), and \( h \) are functions.\n\n- \( h \circ \left( {g \circ f}\right) = \left( {h \circ g}\right) \circ f \), provided the composite functions are defined.\n\n- If \( I \) is defined as \( I\left( x\right) = x \) for all real numbers \( x \), then \( I \circ f = f \circ I = f \) .
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By repeated applications of Definition 5.1, we find \( \left( {h \circ \left( {g \circ f}\right) }\right) \left( x\right) = h\left( {\left( {g \circ f}\right) \left( x\right) }\right) = h\left( {g\left( {f\left( x\right) }\right) }\right) \) . Similarly, \( \left( {\left( {h \circ g}\right) \circ f}\right) \left( x\right) = \left( {h \circ g}\right) \left( {f\left( x\right) }\right) = h\left( {g\left( {f\left( x\right) }\right) }\right) \) . This establishes that the formulas for the two functions are the same. We leave it to the reader to think about why the domains of these two functions are identical, too. These two facts establish the equality \( h \circ \left( {g \circ f}\right) = \left( {h \circ g}\right) \circ f \) . A consequence of the associativity of function composition is that there is no need for parentheses when we write \( h \circ g \circ f \) . The second property can also be verified using Definition 5.1. Recall that the function \( I\left( x\right) = x \) is called the identity function and was introduced in Exercise 73 in Section 2.1. If we compose the function \( I \) with a function \( f \), then we have \( \left( {I \circ f}\right) \left( x\right) = I\left( {f\left( x\right) }\right) = f\left( x\right) \) , and a similar computation shows \( \left( {f \circ I}\right) \left( x\right) = f\left( x\right) \) . This establishes that we have an identity for function composition much in the same way the real number 1 is an identity for real number multiplication. That is, just as for any real number \( x,1 \cdot x = x \cdot 1 = x \), we have for any function \( f, I \circ f = f \circ I = f \) .
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No
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The surface area \( S \) of a sphere is a function of its radius \( r \) and is given by the formula \( S\left( r\right) = {4\pi }{r}^{2} \) . Suppose the sphere is being inflated so that the radius of the sphere is increasing according to the formula \( r\left( t\right) = 3{t}^{2} \), where \( t \) is measured in seconds, \( t \geq 0 \), and \( r \) is measured in inches. Find and interpret \( \left( {S \circ r}\right) \left( t\right) \) .
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Solution. If we look at the functions \( S\left( r\right) \) and \( r\left( t\right) \) individually, we see the former gives the surface area of a sphere of a given radius while the latter gives the radius at a given time. So, given a specific time, \( t \), we could find the radius at that time, \( r\left( t\right) \) and feed that into \( S\left( r\right) \) to find the surface area at that time. From this we see that the surface area \( S \) is ultimately a function of time \( t \) and we find \( \left( {S \circ r}\right) \left( t\right) = S\left( {r\left( t\right) }\right) = {4\pi }{\left( r\left( t\right) \right) }^{2} = {4\pi }{\left( 3{t}^{2}\right) }^{2} = {36\pi }{t}^{4} \) . This formula allows us to compute the surface area directly given the time without going through the ’middle man’ \( r \) .
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Yes
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1. \( F\left( x\right) = \left| {{3x} - 1}\right| \)
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Our goal is to express the function \( F \) as \( F = g \circ f \) for functions \( g \) and \( f \) . From Definition 5.1, we know \( F\left( x\right) = g\left( {f\left( x\right) }\right) \), and we can think of \( f\left( x\right) \) as being the ’inside’ function and \( g \) as being the ’outside’ function. Looking at \( F\left( x\right) = \left| {{3x} - 1}\right| \) from an ’inside versus outside’ perspective, we can think of \( {3x} - 1 \) being inside the absolute value symbols. Taking this cue, we define \( f\left( x\right) = {3x} - 1 \) . At this point, we have \( F\left( x\right) = \left| {f\left( x\right) }\right| \) . What is the outside function? The function which takes the absolute value of its input, \( g\left( x\right) = \left| x\right| \) . Sure enough, \( \left( {g \circ f}\right) \left( x\right) = g\left( {f\left( x\right) }\right) = \left| {f\left( x\right) }\right| = \left| {{3x} - 1}\right| = F\left( x\right) \), so we are done.
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Yes
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Theorem 5.2. Properties of Inverse Functions: Suppose \( f \) and \( g \) are inverse functions.\n\n- The range \( {}^{a} \) of \( f \) is the domain of \( g \) and the domain of \( f \) is the range of \( g \)\n\n- \( f\left( a\right) = b \) if and only if \( g\left( b\right) = a \)\n\n- \( \left( {a, b}\right) \) is on the graph of \( f \) if and only if \( \left( {b, a}\right) \) is on the graph of \( g \)\n\n\( {}^{a} \) Recall this is the set of all outputs of a function.
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Theorem 5.2 is a consequence of Definition 5.2 and the Fundamental Graphing Principle for Functions. We note the third property in Theorem 5.2 tells us that the graphs of inverse functions are reflections about the line \( y = x \) . For a proof of this, see Example 1.1.7 in Section 1.1 and Exercise 72 in Section 2.1.
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No
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Determine if the following functions are one-to-one in two ways: (a) analytically using Definition 5.3 and (b) graphically using the Horizontal Line Test.\n\n1. \( f\left( x\right) = \frac{1 - {2x}}{5} \) 2. \( g\left( x\right) = \frac{2x}{1 - x} \)\n\n3. \( h\left( x\right) = {x}^{2} - {2x} + 4 \) 4. \( F = \{ \left( {-1,1}\right) ,\left( {0,2}\right) ,\left( {2,1}\right) \} \)
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1. (a) To determine if \( f \) is one-to-one analytically, we assume \( f\left( c\right) = f\left( d\right) \) and attempt to deduce that \( c = d \) .\n\n\[ f\left( c\right) = f\left( d\right) \]\n\n\[ \frac{1 - {2c}}{5} = \frac{1 - {2d}}{5} \]\n\n\[ 1 - {2c} = 1 - {2d} \]\n\n\[ - {2c} = - {2d} \]\n\n\[ c = d\checkmark \]\n\nHence, \( f \) is one-to-one.\n\n(b) To check if \( f \) is one-to-one graphically, we look to see if the graph of \( y = f\left( x\right) \) passes the Horizontal Line Test. We have that \( f \) is a non-constant linear function, which means its graph is a non-horizontal line. Thus the graph of \( f \) passes the Horizontal Line Test.
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Yes
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Find the inverse of the following one-to-one functions. Check your answers analytically using function composition and graphically.\n\n1. \( f\left( x\right) = \frac{1 - {2x}}{5} \)
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Solution.\n\n1. As we mentioned earlier, it is possible to think our way through the inverse of \( f \) by recording the steps we apply to \( x \) and the order in which we apply them and then reversing those steps in the reverse order. We encourage the reader to do this. We, on the other hand, will practice the algorithm. We write \( y = f\left( x\right) \) and proceed to switch \( x \) and \( y \)\n\n\[ y = f\left( x\right) \]\n\n\[ y = \frac{1 - {2x}}{5} \]\n\n\[ x = \frac{1 - {2y}}{5}\;\text{ switch }x\text{ and }y \]\n\n\[ {5x} = 1 - {2y} \]\n\n\[ {5x} - 1 = - {2y} \]\n\n\[ \frac{{5x} - 1}{-2} = y \]\n\n\[ y = - \frac{5}{2}x + \frac{1}{2} \]\n\nWe have \( {f}^{-1}\left( x\right) = - \frac{5}{2}x + \frac{1}{2} \) . To check this answer analytically, we first check that \( \left( {{f}^{-1} \circ f}\right) \left( x\right) = \) \( x \) for all \( x \) in the domain of \( f \), which is all real numbers.\n\n\[ \left( {{f}^{-1} \circ f}\right) \left( x\right) = {f}^{-1}\left( {f\left( x\right) }\right) \]\n\n\[ = - \frac{5}{2}f\left( x\right) + \frac{1}{2} \]\n\n\[ = - \frac{5}{2}\left( \frac{1 - {2x}}{5}\right) + \frac{1}{2} \]\n\n\[ = - \frac{1}{2}\left( {1 - {2x}}\right) + \frac{1}{2} \]\n\n\[ = - \frac{1}{2} + x + \frac{1}{2} \]\n\n\[ = x\checkmark \]\n\nWe now check that \( \left( {f \circ {f}^{-1}}\right) \left( x\right) = x \) for all \( x \) in the range of \( f \) which is also all real numbers. (Recall that the domain of \( {f}^{-1} \) ) is the range of \( f \) .)\n\n\[ \left( {f \circ {f}^{-1}}\right) \left( x\right) = f\left( {{f}^{-1}\left( x\right) }\right) \]\n\n\[ = \frac{1 - 2{f}^{-1}\left( x\right) }{5} \]\n\n\[ = \frac{1 - 2\left( {-\frac{5}{2}x + \frac{1}{2}}\right) }{5} \]\n\n\[ = \frac{1 + {5x} - 1}{5} \]\n\n\[ = \frac{5x}{5} \]\n\n\[ = x\sqrt{} \]\n\nTo check our answer graphically, we graph \( y = f\left( x\right) \) and \( y = {f}^{-1}\left( x\right) \) on the same set of axes. \( {}^{5} \)\n\nThey appear to be reflections across the line \( y = x \) .\n\n\( {}^{5} \) Note that if you perform your check on a calculator for more sophisticated functions, you’ll need to take advantage of the 'ZoomSquare' feature to get the correct geometric perspective.
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Yes
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1. \( j\left( x\right) = {x}^{2} - {2x} + 4, x \leq 1 \)
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The function \( j \) is a restriction of the function \( h \) from Example 5.2.1. Since the domain of \( j \) is restricted to \( x \leq 1 \), we are selecting only the ’left half’ of the parabola. We see that the graph of \( j \) passes the Horizontal Line Test and thus \( j \) is invertible.\n\n\n\nWe now use our algorithm \( {}^{7} \) to find \( {j}^{-1}\left( x\right) \).\n\n\[ y = j\left( x\right) \]\n\n\[ y = {x}^{2} - {2x} + 4, x \leq 1 \]\n\n\[ x = {y}^{2} - {2y} + 4, y \leq 1 \]\nswitch \( x \) and \( y \)\n\n\[ 0 = {y}^{2} - {2y} + 4 - x \]\n\n\[ y = \frac{2 \pm \sqrt{{\left( -2\right) }^{2} - 4\left( 1\right) \left( {4 - x}\right) }}{2\left( 1\right) }\text{ quadratic formula,}c = 4 - x \]\n\n\[ y = \frac{2 \pm \sqrt{{4x} - {12}}}{2} \]\n\n\[ y = \frac{2 \pm \sqrt{4\left( {x - 3}\right) }}{2} \]\n\n\[ y = \frac{2 \pm 2\sqrt{x - 3}}{2} \]\n\n\[ y = \frac{2\left( {1 \pm \sqrt{x - 3}}\right) }{2} \]\n\n\[ y = 1 \pm \sqrt{x - 3} \]\n\n\[ y = 1 - \sqrt{x - 3} \]\nsince \( y \leq 1 \).\n\nWe have \( {j}^{-1}\left( x\right) = 1 - \sqrt{x - 3} \). When we simplify \( \left( {{j}^{-1} \circ j}\right) \left( x\right) \), we need to remember that the domain of \( j \) is \( x \leq 1 \).\n\n\[ \left( {{j}^{-1} \circ j}\right) \left( x\right) = {j}^{-1}\left( {j\left( x\right) }\right) \]\n\n\[ = {j}^{-1}\left( {{x}^{2} - {2x} + 4}\right), x \leq 1 \]\n\n\[ = 1 - \sqrt{\left( {{x}^{2} - {2x} + 4}\right) - 3} \]\n\n\[ = 1 - \sqrt{{x}^{2} - {2x} + 1} \]\n\n\[ = 1 - \sqrt{{\left( x - 1\right) }^{2}} \]\n\n\[ = 1 - \left| {x - 1}\right| \]\n\n\[ = 1 - \left( {-\left( {x - 1}\right) }\right) \]\n\n\[ = x\sqrt{} \]\n\nChecking \( j \circ {j}^{-1} \), we get\n\n\[ \left( {j \circ {j}^{-1}}\right) \left( x\right) = j\left( {{j}^{-1}\left( x\right) }\right) \]\n\n\[ = j\left( {1 - \sqrt{x - 3}}\right) \]\n\n\[ = {\left( 1 - \sqrt{x - 3}\right) }^{2} - 2\left( {1 - \sqrt{x - 3}}\right) + 4 \]\n\n\[ = \;1 - 2\sqrt{x - 3} + {\left( \sqrt{x - 3}\right) }^{2} - 2 + 2\sqrt{x - 3} + 4 \]\n\n\[ = 3 + x - 3 \]\n\n\[ = x\sqrt{} \]\n\n\( {}^{7} \) Here, we use the Quadratic Formula to solve for \( y \). For ’completeness,’ we note you can (and should!) also consider solving for \( y \) by ’completing’ the square.\n\nUsing what we know from Section 1.7, we graph \( y = {j}^{-1}\left( x\right) \) and \( y = j\left( x\right) \) below.\n\n
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Yes
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1. Explain why \( p \) is one-to-one and find a formula for \( {p}^{-1}\left( x\right) \). State the restricted domain.
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1. We leave to the reader to show the graph of \( p\left( x\right) = - {1.5x} + {250},0 \leq x \leq {166} \), is a line segment from \( \left( {0,{250}}\right) \) to \( \left( {{166},1}\right) \), and as such passes the Horizontal Line Test. Hence, \( p \) is one-to-one. We find the expression for \( {p}^{-1}\left( x\right) \) as usual and get \( {p}^{-1}\left( x\right) = \frac{{500} - {2x}}{3} \). The domain of \( {p}^{-1} \) should match the range of \( p \), which is \( \left\lbrack {1,{250}}\right\rbrack \), and as such, we restrict the domain of \( {p}^{-1} \) to \( 1 \leq x \leq {250} \).
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No
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Theorem 5.6. Properties of Radicals: Let \( x \) and \( y \) be real numbers and \( m \) and \( n \) be natural numbers. If \( \sqrt[n]{x},\sqrt[n]{y} \) are real numbers, then\n\n- Product Rule: \( \sqrt[n]{xy} = \sqrt[n]{x}\sqrt[n]{y} \)\n\n- Powers of Radicals: \( \sqrt[n]{{x}^{m}} = {\left( \sqrt[n]{x}\right) }^{m} \)\n\n- Quotient Rule: \( \sqrt[n]{\frac{x}{y}} = \frac{\sqrt[n]{x}}{\sqrt[n]{y}} \), provided \( y \neq 0 \) .\n\n- If \( n \) is odd, \( \sqrt[n]{{x}^{n}} = x \) ; if \( n \) is even, \( \sqrt[n]{{x}^{n}} = \left| x\right| \) .
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The proof of Theorem 5.6 is based on the definition of the principal roots and properties of exponents. To establish the product rule, consider the following. If \( n \) is odd, then by definition \( \sqrt[n]{xy} \) is the unique real number such that \( {\left( \sqrt[n]{xy}\right) }^{n} = {xy} \) . Given that \( {\left( \sqrt[n]{x}\sqrt[n]{y}\right) }^{n} = {\left( \sqrt[n]{x}\right) }^{n}{\left( \sqrt[n]{y}\right) }^{n} = {xy} \) , it must be the case that \( \sqrt[n]{xy} = \sqrt[n]{x}\sqrt[n]{y} \) . If \( n \) is even, then \( \sqrt[n]{xy} \) is the unique non-negative real number such that \( {\left( \sqrt[n]{xy}\right) }^{n} = {xy} \) . Also note that since \( n \) is even, \( \sqrt[n]{x} \) and \( \sqrt[n]{y} \) are also non-negative and hence so is \( \sqrt[n]{x}\sqrt[n]{y} \) . Proceeding as above, we find that \( \sqrt[n]{xy} = \sqrt[n]{x}\sqrt[n]{y} \) . The quotient rule is proved similarly and is left as an exercise. The power rule results from repeated application of the product rule, so long as \( \sqrt[n]{x} \) is a real number to start with. \( {}^{2} \) The last property is an application of the power rule when \( n \) is odd, and the occurrence of the absolute value when \( n \) is even is due to the requirement that \( \sqrt[n]{x} \geq 0 \) in Definition 5.4. For instance, \( \sqrt[4]{{\left( -2\right) }^{4}} = \sqrt[4]{16} = 2 = \left| {-2}\right| \), not -2 . It's this last property which makes compositions of roots and powers delicate. This is especially true when we use exponential notation for radicals. Recall the following definition.
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No
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For the following functions, state their domains and create sign diagrams. Check your answer graphically using your calculator.
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1. As far as domain is concerned, \( f\left( x\right) \) has no denominators and no even roots, which means its domain is \( \left( {-\infty ,\infty }\right) \) . To create the sign diagram, we find the zeros of \( f \) .\n\n\[ f\left( x\right) = 0 \]\n\n\[ {3x}\sqrt[3]{2 - x} = 0 \]\n\n\[ {3x} = 0\text{ or }\sqrt[3]{2 - x} = 0 \]\n\n\[ x = 0\text{ or }{\left( \sqrt[3]{2 - x}\right) }^{3} = {0}^{3} \]\n\n\[ x = 0\text{ or }2 - x = 0 \]\n\n\[ x = 0\text{ or }x = 2 \]\n\nThe zeros 0 and 2 divide the real number line into three test intervals. The sign diagram and accompanying graph are below. Note that the intervals on which \( f \) is \( \left( +\right) \) correspond to where the graph of \( f \) is above the \( x \) -axis, and where the graph of \( f \) is below the \( x \) -axis we have that \( f \) is \( \left( -\right) \) . The calculator suggests something mysterious happens near \( x = 2 \) . Zooming in shows the graph becomes nearly vertical there. You'll have to wait until Calculus to fully understand this phenomenon.\n\n\[ \begin{matrix} \left( -\right) 0 & \left( +\right) & 0 & \left( -\right) \\ + & 0 & & + \\ & 0 & & 2 \end{matrix} \]\n\n\[ y = f\left( x\right) \;y = f\left( x\right) \text{ near }x = 2. \]
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Yes
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1. Express the total cost \( C \) of connecting the Junction Box to the Outpost as a function of \( x \) , the number of miles the cable is run along Route 117 before heading off road directly towards the Outpost. Determine a reasonable applied domain for the problem.
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1. The cost is broken into two parts: the cost to run cable along Route 117 at \( \$ {15} \) per mile, and the cost to run it off road at \( \$ {20} \) per mile. Since \( x \) represents the miles of cable run along Route 117, the cost for that portion is \( {15x} \) . From the diagram, we see that the number of miles the cable is run off road is \( z \), so the cost of that portion is \( {20z} \) . Hence, the total cost is \( C = {15x} + {20z} \) . Our next goal is to determine \( z \) as a function of \( x \) . The diagram suggests we can use the Pythagorean Theorem to get \( {y}^{2} + {30}^{2} = {z}^{2} \) . But we also see \( x + y = {50} \) so that \( y = {50} - x \) . Hence, \( {z}^{2} = {\left( {50} - x\right) }^{2} + {900} \) . Solving for \( z \), we obtain \( z = \pm \sqrt{{\left( {50} - x\right) }^{2} + {900}} \) . Since \( z \) represents a distance, we choose \( z = \sqrt{{\left( {50} - x\right) }^{2} + {900}} \) so that our cost as a function of \( x \) only is given by\n\n\[ C\left( x\right) = {15x} + {20}\sqrt{{\left( {50} - x\right) }^{2} + {900}} \]\n\nFrom the context of the problem, we have \( 0 \leq x \leq {50} \) .
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Yes
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1. Find and interpret \( V\left( 0\right) \) .
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To find \( V\left( 0\right) \), we replace \( x \) with 0 to obtain \( V\left( 0\right) = {25}{\left( \frac{4}{5}\right) }^{0} = {25} \) . Since \( x \) represents the age of the car in years, \( x = 0 \) corresponds to the car being brand new. Since \( V\left( x\right) \) is measured in thousands of dollars, \( V\left( 0\right) = {25} \) corresponds to a value of \( \$ {25},{000} \) . Putting it all together, we interpret \( V\left( 0\right) = {25} \) to mean the purchase price of the car was \( \$ {25},{000} \) .
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Yes
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1. Find and interpret \( T\left( 0\right) \) .
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To find \( T\left( 0\right) \), we replace every occurrence of the independent variable \( t \) with 0 to obtain \( T\left( 0\right) = {70} + {90}{e}^{-{0.1}\left( 0\right) } = {160} \) . This means that the coffee was served at \( {160}^{ \circ }\mathrm{F} \) .
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Yes
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Theorem 6.2. Properties of Logarithmic Functions: Suppose \( f\left( x\right) = {\log }_{b}\left( x\right) \) .
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- The domain of \( f \) is \( \left( {0,\infty }\right) \) and the range of \( f \) is \( \left( {-\infty ,\infty }\right) \) .\n\n- \( \left( {1,0}\right) \) is on the graph of \( f \) and \( x = 0 \) is a vertical asymptote of the graph of \( f \) .\n\n- \( f \) is one-to-one, continuous and smooth\n\n- \( {b}^{a} = c \) if and only if \( {\log }_{b}\left( c\right) = a \) . That is, \( {\log }_{b}\left( c\right) \) is the exponent you put on \( b \) to obtain \( c \) .\n\n- \( {\log }_{b}\left( {b}^{x}\right) = x \) for all \( x \) and \( {b}^{{\log }_{b}\left( x\right) } = x \) for all \( x > 0 \)\n\n- If \( b > 1 \) : - If \( 0 < b < 1 \) :\n\n- \( f \) is always increasing - \( f \) is always decreasing\n\n- As \( x \rightarrow {0}^{ + }, f\left( x\right) \rightarrow - \infty \) - As \( x \rightarrow {0}^{ + }, f\left( x\right) \rightarrow \infty \)\n\n- As \( x \rightarrow \infty, f\left( x\right) \rightarrow \infty \) - As \( x \rightarrow \infty, f\left( x\right) \rightarrow - \infty \)\n\n- The graph of \( f \) resembles: - The graph of \( f \) resembles:\n\n\n\n\( y = {\log }_{b}\left( x\right), b > 1 \)\n\n\n\n\( y = {\log }_{b}\left( x\right) ,0 < b < 1 \)
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Yes
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Simplify the following.\n\n1. \( {\log }_{3}\left( {81}\right) \) 2. \( {\log }_{2}\left( \frac{1}{8}\right) \) 3. \( {\log }_{\sqrt{5}}\left( {25}\right) \) 4. \( \ln \left( \sqrt[3]{{e}^{2}}\right) \)\n\n5. \( \log \left( {0.001}\right) \) 6. \( {2}^{{\log }_{2}\left( 8\right) } \) 7. \( {117}^{-{\log }_{117}\left( 6\right) } \)
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Solution.\n\n1. The number \( {\log }_{3}\left( {81}\right) \) is the exponent we put on 3 to get 81 . As such, we want to write 81 as a power of 3 . We find \( {81} = {3}^{4} \), so that \( {\log }_{3}\left( {81}\right) = 4 \) .\n\n2. To find \( {\log }_{2}\left( \frac{1}{8}\right) \), we need rewrite \( \frac{1}{8} \) as a power of 2 . We find \( \frac{1}{8} = \frac{1}{{2}^{3}} = {2}^{-3} \), so \( {\log }_{2}\left( \frac{1}{8}\right) = - 3 \) .\n\n3. To determine \( {\log }_{\sqrt{5}}\left( {25}\right) \), we need to express 25 as a power of \( \sqrt{5} \) . We know \( {25} = {5}^{2} \), and \( 5 = {\left( \sqrt{5}\right) }^{2} \), so we have \( {25} = {\left( {\left( \sqrt{5}\right) }^{2}\right) }^{2} = {\left( \sqrt{5}\right) }^{4} \) . We get \( {\log }_{\sqrt{5}}\left( {25}\right) = 4 \) .\n\n4. First, recall that the notation \( \ln \left( \sqrt[3]{{e}^{2}}\right) \) means \( {\log }_{e}\left( \sqrt[3]{{e}^{2}}\right) \), so we are looking for the exponent to put on \( e \) to obtain \( \sqrt[3]{{e}^{2}} \) . Rewriting \( \sqrt[3]{{e}^{2}} = {e}^{2/3} \), we find \( \ln \left( \sqrt[3]{{e}^{2}}\right) = \ln \left( {e}^{2/3}\right) = \frac{2}{3} \) .\n\n5. Rewriting \( \log \left( {0.001}\right) \) as \( {\log }_{10}\left( {0.001}\right) \), we see that we need to write 0.001 as a power of 10 . We have \( {0.001} = \frac{1}{1000} = \frac{1}{{10}^{3}} = {10}^{-3} \) . Hence, \( \log \left( {0.001}\right) = \log \left( {10}^{-3}\right) = - 3 \) .\n\n6. We can use Theorem 6.2 directly to simplify \( {2}^{{\log }_{2}\left( 8\right) } = 8 \) . We can also understand this problem by first finding \( {\log }_{2}\left( 8\right) \) . By definition, \( {\log }_{2}\left( 8\right) \) is the exponent we put on 2 to get 8 . Since \( 8 = {2}^{3} \), we have \( {\log }_{2}\left( 8\right) = 3 \) . We now substitute to find \( {2}^{{\log }_{2}\left( 8\right) } = {2}^{3} = 8 \) .\n\n7. From Theorem 6.2, we know \( {117}^{{\log }_{117}\left( 6\right) } = 6 \), but we cannot directly apply this formula to the expression \( {117}^{-{\log }_{117}\left( 6\right) } \) . (Can you see why?) At this point, we use a property of exponents followed by Theorem 6.2 to \( {\text{get}}^{9} \)\n\n\[ \n{117}^{-{\log }_{117}\left( 6\right) } = \frac{1}{{117}^{{\log }_{117}\left( 6\right) }} = \frac{1}{6} \n\]\n\n\( {}^{9} \) It is worth a moment of your time to think your way through why \( {117}^{{\log }_{117}\left( 6\right) } = 6 \) . By definition, \( {\log }_{117}\left( 6\right) \) is the exponent we put on 117 to get 6 . What are we doing with this exponent? We are putting it on 117. By definition we get 6 . In other words, the exponential function \( f\left( x\right) = {117}^{x} \) undoes the logarithmic function \( g\left( x\right) = {\log }_{117}\left( x\right) \) .
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Yes
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Find the domain of the following functions. Check your answers graphically using the calculator.\n\n1. \( f\left( x\right) = 2\log \left( {3 - x}\right) - 1 \)
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## Solution.\n\n1. We set \( 3 - x > 0 \) to obtain \( x < 3 \), or \( \left( {-\infty ,3}\right) \) . The graph from the calculator below verifies this. Note that we could have graphed \( f \) using transformations. Taking a cue from Theorem 1.7, we rewrite \( f\left( x\right) = 2{\log }_{10}\left( {-x + 3}\right) - 1 \) and find the main function involved is \( y = h\left( x\right) = {\log }_{10}\left( x\right) \) . We select three points to track, \( \left( {\frac{1}{10}, - 1}\right) ,\left( {1,0}\right) \) and \( \left( {{10},1}\right) \), along with the vertical asymptote \( x = 0 \) . Since \( f\left( x\right) = {2h}\left( {-x + 3}\right) - 1 \), Theorem 1.7 tells us that to obtain the destinations of these points, we first subtract 3 from the \( x \) -coordinates (shifting the graph left 3 units), then divide (multiply) by the \( x \) -coordinates by -1 (causing a reflection across the \( y \) -axis). These transformations apply to the vertical asymptote \( x = 0 \) as well. Subtracting 3 gives us \( x = - 3 \) as our asymptote, then multplying by -1 gives us the vertical asymptote \( x = 3 \) . Next, we multiply the \( y \) -coordinates by 2 which results in a vertical stretch by a factor of 2, then we finish by subtracting 1 from the \( y \) -coordinates which shifts the graph down 1 unit. We leave it to the reader to perform the indicated arithmetic on the points themselves and to verify the graph produced by the calculator below.
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Yes
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1. Graph \( f \) using transformations and state the domain and range of \( f \) .
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If we identify \( g\left( x\right) = {2}^{x} \), we see \( f\left( x\right) = g\left( {x - 1}\right) - 3 \) . We pick the points \( \left( {-1,\frac{1}{2}}\right) ,\left( {0,1}\right) \) and \( \left( {1,2}\right) \) on the graph of \( g \) along with the horizontal asymptote \( y = 0 \) to track through the transformations. By Theorem 1.7 we first add 1 to the \( x \) -coordinates of the points on the graph of \( g \) (shifting \( g \) to the right 1 unit) to get \( \left( {0,\frac{1}{2}}\right) ,\left( {1,1}\right) \) and \( \left( {2,2}\right) \) . The horizontal asymptote remains \( y = 0 \) . Next, we subtract 3 from the \( y \) -coordinates, shifting the graph down 3 units. We get the points \( \left( {0, - \frac{5}{2}}\right) ,\left( {1, - 2}\right) \) and \( \left( {2, - 1}\right) \) with the horizontal asymptote now at \( y = - 3 \) . Connecting the dots in the order and manner as they were on the graph of \( g \), we get the graph below. We see that the domain of \( f \) is the same as \( g \), namely \( \left( {-\infty ,\infty }\right) \) , but that the range of \( f \) is \( \left( {-3,\infty }\right) \) .
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Yes
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Theorem 6.6. (Algebraic Properties of Logarithmic Functions) Let \( g\left( x\right) = {\log }_{b}\left( x\right) \) be a logarithmic function \( \left( {b > 0, b \neq 1}\right) \) and let \( u > 0 \) and \( w > 0 \) be real numbers.\n\n- Product Rule: \( g\left( {uw}\right) = g\left( u\right) + g\left( w\right) \) . In other words, \( {\log }_{b}\left( {uw}\right) = {\log }_{b}\left( u\right) + {\log }_{b}\left( w\right) \)
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Consider the product rule: \( {\log }_{b}\left( {uw}\right) = {\log }_{b}\left( u\right) + {\log }_{b}\left( w\right) \) . Let \( a = {\log }_{b}\left( {uw}\right), c = {\log }_{b}\left( u\right) \), and \( d = {\log }_{b}\left( w\right) \) . Then, by definition, \( {b}^{a} = {uw},{b}^{c} = u \) and \( {b}^{d} = w \) . Hence, \( {b}^{a} = {uw} = {b}^{c}{b}^{d} = {b}^{c + d} \), so that \( {b}^{a} = {b}^{c + d} \) . By the one-to-one property of \( {b}^{x} \), we have \( a = c + d \) . In other words, \( {\log }_{b}\left( {uw}\right) = {\log }_{b}\left( u\right) + {\log }_{b}\left( w\right) \).
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Yes
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Expand the following using the properties of logarithms and simplify. Assume when necessary that all quantities represent positive real numbers.\n\n1. \( {\log }_{2}\left( \frac{8}{x}\right) \)
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To expand \( {\log }_{2}\left( \frac{8}{x}\right) \), we use the Quotient Rule identifying \( u = 8 \) and \( w = x \) and simplify.\n\n\[ {\log }_{2}\left( \frac{8}{x}\right) = {\log }_{2}\left( 8\right) - {\log }_{2}\left( x\right) \;\text{ Quotient Rule } \]\n\n\[ = 3 - {\log }_{2}\left( x\right) \;\text{Since}{2}^{3} = 8 \]\n\n\[ = - {\log }_{2}\left( x\right) + 3 \]
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Yes
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Use the properties of logarithms to write the following as a single logarithm.
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1. The difference of logarithms requires the Quotient Rule: \( {\log }_{3}\left( {x - 1}\right) - {\log }_{3}\left( {x + 1}\right) = {\log }_{3}\left( \frac{x - 1}{x + 1}\right) \) .
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Yes
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Theorem 6.7. (Change of Base Formulas) Let \( a, b > 0, a, b \neq 1 \) .\n\n- \( {a}^{x} = {b}^{x{\log }_{b}\left( a\right) } \) for all real numbers \( x \) .\n\n- \( {\log }_{a}\left( x\right) = \frac{{\log }_{b}\left( x\right) }{{\log }_{b}\left( a\right) } \) for all real numbers \( x > 0 \) .
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The proofs of the Change of Base formulas are a result of the other properties studied in this section. If we start with \( {b}^{x{\log }_{b}\left( a\right) } \) and use the Power Rule in the exponent to rewrite \( x{\log }_{b}\left( a\right) \) as \( {\log }_{b}\left( {a}^{x}\right) \) and then apply one of the Inverse Properties in Theorem 6.3, we get\n\n\[ {b}^{x{\log }_{b}\left( a\right) } = {b}^{{\log }_{b}\left( {a}^{x}\right) } = {a}^{x}, \]\n\nas required. To verify the logarithmic form of the property, we also use the Power Rule and an Inverse Property. We note that\n\n\[ {\log }_{a}\left( x\right) \cdot {\log }_{b}\left( a\right) = {\log }_{b}\left( {a}^{{\log }_{a}\left( x\right) }\right) = {\log }_{b}\left( x\right) , \]\n\nand we get the result by dividing through by \( {\log }_{b}\left( a\right) \) .
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Yes
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Example 6.2.3. Use an appropriate change of base formula to convert the following expressions to ones with the indicated base. Verify your answers using a calculator, as appropriate.\n\n1. \( {3}^{2} \) to base 10 2. \( {2}^{x} \) to base \( e \)
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## Solution.\n\n1. We apply the Change of Base formula with \( a = 3 \) and \( b = {10} \) to obtain \( {3}^{2} = {10}^{2\log \left( 3\right) } \) . Typing the latter in the calculator produces an answer of 9 as required.\n\n2. Here, \( a = 2 \) and \( b = e \) so we have \( {2}^{x} = {e}^{x\ln \left( 2\right) } \) . To verify this on our calculator, we can graph \( f\left( x\right) = {2}^{x} \) and \( g\left( x\right) = {e}^{x\ln \left( 2\right) } \) . Their graphs are indistinguishable which provides evidence that they are the same function.
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Yes
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1. \( {2}^{3x} = {16}^{1 - x} \)
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Since 16 is a power of 2, we can rewrite \( {2}^{3x} = {16}^{1 - x} \) as \( {2}^{3x} = {\left( {2}^{4}\right) }^{1 - x} \) . Using properties of exponents, we get \( {2}^{3x} = {2}^{4\left( {1 - x}\right) } \) . Using the one-to-one property of exponential functions, we get \( {3x} = 4\left( {1 - x}\right) \) which gives \( x = \frac{4}{7} \) . To check graphically, we set \( f\left( x\right) = {2}^{3x} \) and \( g\left( x\right) = {16}^{1 - x} \) and see that they intersect at \( x = \frac{4}{7} \approx {0.5714} \).
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Yes
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1. \( {2}^{{x}^{2} - {3x}} - {16} \geq 0 \)
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Since we already have 0 on one side of the inequality, we set \( r\left( x\right) = {2}^{{x}^{2} - {3x}} - {16} \) . The domain of \( r \) is all real numbers, so in order to construct our sign diagram, we need to find the zeros of \( r \) . Setting \( r\left( x\right) = 0 \) gives \( {2}^{{x}^{2} - {3x}} - {16} = 0 \) or \( {2}^{{x}^{2} - {3x}} = {16} \) . Since \( {16} = {2}^{4} \) we have \( {2}^{{x}^{2} - {3x}} = {2}^{4} \) , so by the one-to-one property of exponential functions, \( {x}^{2} - {3x} = 4 \) . Solving \( {x}^{2} - {3x} - 4 = 0 \) gives \( x = 4 \) and \( x = - 1 \) . From the sign diagram, we see \( r\left( x\right) \geq 0 \) on \( \left( {-\infty , - 1\rbrack \cup \lbrack 4,\infty }\right) \), which corresponds to where the graph of \( y = r\left( x\right) = {2}^{{x}^{2} - {3x}} - {16} \), is on or above the \( x \) -axis.\n\n\[ \begin{matrix} \left( +\right) 0 & \left( -\right) & 0 & \left( +\right) \\ + & + & + & \rightarrow \\ & - 1 & 4 & \end{matrix} \]\n\n\[ y = r\left( x\right) = {2}^{{x}^{2} - {3x}} - {16} \]
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Yes
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Recall from Example 6.1.2 that the temperature of coffee \( T \) (in degrees Fahrenheit) \( t \) minutes after it is served can be modeled by \( T\left( t\right) = {70} + {90}{e}^{-{0.1t}} \) . When will the coffee be warmer than \( {100}^{ \circ }\mathrm{F} \) ?
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Solution. We need to find when \( T\left( t\right) > {100} \), or in other words, we need to solve the inequality \( {70} + {90}{e}^{-{0.1t}} > {100} \) . Getting 0 on one side of the inequality, we have \( {90}{e}^{-{0.1t}} - {30} > 0 \), and we set \( r\left( t\right) = {90}{e}^{-{0.1t}} - {30} \) . The domain of \( r \) is artificially restricted due to the context of the problem to \( \lbrack 0,\infty ) \), so we proceed to find the zeros of \( r \) . Solving \( {90}{e}^{-{0.1t}} - {30} = 0 \) results in \( {e}^{-{0.1t}} = \frac{1}{3} \) so that \( t = - {10}\ln \left( \frac{1}{3}\right) \) which, after a quick application of the Power Rule leaves us with \( t = {10}\ln \left( 3\right) \) . If we wish to avoid using the calculator to choose test values, we note that since \( 1 < 3 \) , \( 0 = \ln \left( 1\right) < \ln \left( 3\right) \) so that \( {10}\ln \left( 3\right) > 0 \) . So we choose \( t = 0 \) as a test value in \( \lbrack 0,{10}\ln \left( 3\right) ) \) . Since \( 3 < 4,{10}\ln \left( 3\right) < {10}\ln \left( 4\right) \), so the latter is our choice of a test value for the interval \( \left( {{10}\ln \left( 3\right) ,\infty }\right) \) . Our sign diagram is below, and next to it is our graph of \( y = T\left( t\right) \) from Example 6.1.2 with the horizontal line \( y = {100} \) . \n\nIn order to interpret what this means in the context of the real world, we need a reasonable approximation of the number \( {10}\ln \left( 3\right) \approx {10.986} \) . This means it takes approximately 11 minutes for the coffee to cool to \( {100}^{ \circ }\mathrm{F} \) . Until then, the coffee is warmer than that. \( {}^{8} \)
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Yes
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The function \( f\left( x\right) = \frac{5{e}^{x}}{{e}^{x} + 1} \) is one-to-one. Find a formula for \( {f}^{-1}\left( x\right) \) and check your answer graphically using your calculator.
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Solution. We start by writing \( y = f\left( x\right) \), and interchange the roles of \( x \) and \( y \) . To solve for \( y \), we first clear denominators and then isolate the exponential function.\n\n\[ y = \frac{5{e}^{x}}{{e}^{x} + 1} \]\n\n\[ x = \frac{5{e}^{y}}{{e}^{y} + 1}\;\text{ Switch }x\text{ and }y \]\n\n\[ x\left( {{e}^{y} + 1}\right) = 5{e}^{y} \]\n\n\[ x{e}^{y} + x = 5{e}^{y} \]\n\n\[ x = 5{e}^{y} - x{e}^{y} \]\n\n\[ x = {e}^{y}\left( {5 - x}\right) \]\n\n\[ {e}^{y} = \frac{x}{5 - x} \]\n\n\[ \ln \left( {e}^{y}\right) = \ln \left( \frac{x}{5 - x}\right) \]\n\n\[ y = \ln \left( \frac{x}{5 - x}\right) \]\n\nWe claim \( {f}^{-1}\left( x\right) = \ln \left( \frac{x}{5 - x}\right) \) . To verify this analytically, we would need to verify the compositions \( \left( {{f}^{-1} \circ f}\right) \left( x\right) = x \) for all \( x \) in the domain of \( f \) and that \( \left( {f \circ {f}^{-1}}\right) \left( x\right) = x \) for all \( x \) in the domain of \( {f}^{-1} \) . We leave this to the reader. To verify our solution graphically, we graph \( y = f\left( x\right) = \frac{5{e}^{x}}{{e}^{x} + 1} \) and \( y = g\left( x\right) = \ln \left( \frac{x}{5 - x}\right) \) on the same set of axes and observe the symmetry about the line \( y = x \) . Note the domain of \( f \) is the range of \( g \) and vice-versa.
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No
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1. \( {\log }_{117}\left( {1 - {3x}}\right) = {\log }_{117}\left( {{x}^{2} - 3}\right) \)
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Since we have the same base on both sides of the equation \( {\log }_{117}\left( {1 - {3x}}\right) = {\log }_{117}\left( {{x}^{2} - 3}\right) \) , we equate what’s inside the logs to get \( 1 - {3x} = {x}^{2} - 3 \) . Solving \( {x}^{2} + {3x} - 4 = 0 \) gives \( x = - 4 \) and \( x = 1 \) . To check these answers using the calculator, we make use of the change of base formula and graph \( f\left( x\right) = \frac{\ln \left( {1 - {3x}}\right) }{\ln \left( {117}\right) } \) and \( g\left( x\right) = \frac{\ln \left( {{x}^{2} - 3}\right) }{\ln \left( {117}\right) } \) and we see they intersect only at \( x = - 4 \) . To see what happened to the solution \( x = 1 \), we substitute it into our original equation to obtain \( {\log }_{117}\left( {-2}\right) = {\log }_{117}\left( {-2}\right) \) . While these expressions look identical, neither is a real number, \( {}^{1} \) which means \( x = 1 \) is not in the domain of the original equation, and is not a solution.
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Yes
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1. \( \frac{1}{\ln \left( x\right) + 1} \leq 1 \)
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We start solving \( \frac{1}{\ln \left( x\right) + 1} \leq 1 \) by getting 0 on one side of the inequality: \( \frac{1}{\ln \left( x\right) + 1} - 1 \leq 0 \) . Getting a common denominator yields \( \frac{1}{\ln \left( x\right) + 1} - \frac{\ln \left( x\right) + 1}{\ln \left( x\right) + 1} \leq 0 \) which reduces to \( \frac{-\ln \left( x\right) }{\ln \left( x\right) + 1} \leq 0 \) , or \( \frac{\ln \left( x\right) }{\ln \left( x\right) + 1} \geq 0 \) . We define \( r\left( x\right) = \frac{\ln \left( x\right) }{\ln \left( x\right) + 1} \) and set about finding the domain and the zeros of \( r \) . Due to the appearance of the term \( \ln \left( x\right) \), we require \( x > 0 \) . In order to keep the denominator away from zero, we solve \( \ln \left( x\right) + 1 = 0 \) so \( \ln \left( x\right) = - 1 \), so \( x = {e}^{-1} = \frac{1}{e} \) . Hence, the domain of \( r \) is \( \left( {0,\frac{1}{e}}\right) \cup \left( {\frac{1}{e},\infty }\right) \) . To find the zeros of \( r \), we set \( r\left( x\right) = \frac{\ln \left( x\right) }{\ln \left( x\right) + 1} = 0 \) so that \( \ln \left( x\right) = 0 \), and we find \( x = {e}^{0} = 1 \) . In order to determine test values for \( r \) without resorting to the calculator, we need to find numbers between \( 0,\frac{1}{e} \), and 1 which have a base of \( e \) . Since \( e \approx {2.718} > 1,0 < \frac{1}{{e}^{2}} < \frac{1}{e} < \frac{1}{\sqrt{e}} < 1 < e \) . To determine the sign of \( r\left( \frac{1}{{e}^{2}}\right) \), we use the fact that \( \ln \left( \frac{1}{{e}^{2}}\right) = \ln \left( {e}^{-2}\right) = - 2 \), and find \( r\left( \frac{1}{{e}^{2}}\right) = \frac{-2}{-2 + 1} = 2 \), which is \( \left( +\right) \) . The rest of the test values are determined similarly. From our sign diagram, we find the solution to be \( \left( {0,\frac{1}{e}}\right) \cup \lbrack 1,\infty ) \) . Graphing \( f\left( x\right) = \frac{1}{\ln \left( x\right) + 1} \) and \( g\left( x\right) = 1 \), we see the graph of \( f \) is below the graph of \( g \) on the solution intervals, and that the graphs intersect at \( x = 1 \) .
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Yes
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In order to successfully breed Ippizuti fish the \( \mathrm{{pH}} \) of a freshwater tank must be at least 7.8 but can be no more than 8.5. Determine the corresponding range of hydrogen ion concentration, and check your answer using a calculator.
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Solution. Recall from Exercise 77 in Section 6.1 that \( \mathrm{{pH}} = - \log \left\lbrack {\mathrm{H}}^{ + }\right\rbrack \) where \( \left\lbrack {\mathrm{H}}^{ + }\right\rbrack \) is the hydrogen ion concentration in moles per liter. We require \( {7.8} \leq - \log \left\lbrack {\mathrm{H}}^{ + }\right\rbrack \leq {8.5} \) or \( - {7.8} \geq \log \left\lbrack {\mathrm{H}}^{ + }\right\rbrack \geq - {8.5} \) . To solve this compound inequality we solve \( - {7.8} \geq \log \left\lbrack {\mathrm{H}}^{ + }\right\rbrack \) and \( \log \left\lbrack {\mathrm{H}}^{ + }\right\rbrack \geq - {8.5} \) and take the intersection of the solution sets. \( {}^{3} \) The former inequality yields \( 0 < \left\lbrack {\mathrm{H}}^{ + }\right\rbrack \leq {10}^{-{7.8}} \) and the latter yields \( \left\lbrack {\mathrm{H}}^{ + }\right\rbrack \geq {10}^{-{8.5}} \) . Taking the intersection gives us our final answer \( {10}^{-{8.5}} \leq \left\lbrack {\mathrm{H}}^{ + }\right\rbrack \leq {10}^{-{7.8}} \) . (Your Chemistry professor may want the answer written as \( {3.16} \times {10}^{-9} \leq \left\lbrack {\mathrm{H}}^{ + }\right\rbrack \leq {1.58} \times {10}^{-8} \) .) After carefully adjusting the viewing window on the graphing calculator we see that the graph of \( f\left( x\right) = - \log \left( x\right) \) lies between the lines \( y = {7.8} \) and \( y = {8.5} \) on the interval \( \left\lbrack {{3.16} \times {10}^{-9},{1.58} \times {10}^{-8}}\right\rbrack \) .
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Yes
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The function \( f\left( x\right) = \frac{\log \left( x\right) }{1 - \log \left( x\right) } \) is one-to-one. Find a formula for \( {f}^{-1}\left( x\right) \) and check your answer graphically using your calculator.
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Solution. We first write \( y = f\left( x\right) \) then interchange the \( x \) and \( y \) and solve for \( y \) .\n\n\[ y = f\left( x\right) \]\n\n\[ y = \frac{\log \left( x\right) }{1 - \log \left( x\right) } \]\n\n\[ x = \frac{\log \left( y\right) }{1 - \log \left( y\right) } \]\nInterchange \( x \) and \( y \) .\n\n\[ x\left( {1 - \log \left( y\right) }\right) = \log \left( y\right) \]\n\n\[ x - x\log \left( y\right) = \log \left( y\right) \]\n\n\[ x = x\log \left( y\right) + \log \left( y\right) \]\n\n\[ x = \left( {x + 1}\right) \log \left( y\right) \]\n\n\[ \frac{x}{x + 1} = \log \left( y\right) \]\n\n\[ y = {10}^{\frac{x}{x + 1}}\;\text{Rewrite as an exponential equation.} \]\n\nWe have \( {f}^{-1}\left( x\right) = {10}^{\frac{x}{x + 1}} \) . Graphing \( f \) and \( {f}^{-1} \) on the same viewing window yields\n\n\n\n\( y = f\left( x\right) = \frac{\log \left( x\right) }{1 - \log \left( x\right) } \) and \( \mathbf{y} = \mathbf{g}\left( \mathbf{x}\right) = {\mathbf{{10}}}^{\frac{\mathbf{x}}{\mathbf{x} + \mathbf{1}}} \)
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Yes
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1. Express the amount \( A \) in the account as a function of the term of the investment \( t \) in years.
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1. Substituting \( P = {2000}, r = {0.07125} \), and \( n = {12} \) (since interest is compounded monthly) into Equation 6.2 yields \( A\left( t\right) = {2000}{\left( 1 + \frac{0.07125}{12}\right) }^{12t} = {2000}{\left( {1.0059375}\right) }^{12t} \) .
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Yes
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In order to perform arthrosclerosis research, epithelial cells are harvested from discarded umbilical tissue and grown in the laboratory. A technician observes that a culture of twelve thousand cells grows to five million cells in one week. Assuming that the cells follow The Law of Uninhibited Growth, find a formula for the number of cells, \( N \), in thousands, after \( t \) days.
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Solution. We begin with \( N\left( t\right) = {N}_{0}{e}^{kt} \) . Since \( N \) is to give the number of cells in thousands, we have \( {N}_{0} = {12} \), so \( N\left( t\right) = {12}{e}^{kt} \) . In order to complete the formula, we need to determine the growth rate \( k \) . We know that after one week, the number of cells has grown to five million. Since \( t \) measures days and the units of \( N \) are in thousands, this translates mathematically to \( N\left( 7\right) = {5000} \) . We get the equation \( {12}{e}^{7k} = {5000} \) which gives \( k = \frac{1}{7}\ln \left( \frac{1250}{3}\right) \) . Hence, \( N\left( t\right) = {12}{e}^{\frac{t}{7}\ln \left( \frac{1250}{3}\right) } \) . Of course, in practice, we would approximate \( k \) to some desired accuracy, say \( k \approx {0.8618} \), which we can interpret as an \( {86.18}\% \) daily growth rate for the cells.
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Yes
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Iodine-131 is a commonly used radioactive isotope used to help detect how well the thyroid is functioning. Suppose the decay of Iodine-131 follows the model given in Equation 6.5, and that the half-life \( {}^{10} \) of Iodine-131 is approximately 8 days. If 5 grams of Iodine-131 is present initially, find a function which gives the amount of Iodine-131, \( A \), in grams, \( t \) days later.
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Solution. Since we start with 5 grams initially, Equation 6.5 gives \( A\left( t\right) = 5{e}^{kt} \) . Since the half-life is 8 days, it takes 8 days for half of the Iodine-131 to decay, leaving half of it behind. Hence, \( A\left( 8\right) = {2.5} \) which means \( 5{e}^{8k} = {2.5} \) . Solving, we get \( k = \frac{1}{8}\ln \left( \frac{1}{2}\right) = - \frac{\ln \left( 2\right) }{8} \approx - {0.08664} \), which we can interpret as a loss of material at a rate of \( {8.664}\% \) daily. Hence, \( A\left( t\right) = 5{e}^{-\frac{t\ln \left( 2\right) }{8}} \approx 5{e}^{-{0.08664t}} \) .
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Yes
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1. Assuming the temperature of the roast follows Newton's Law of Warming, find a formula for the temperature of the roast \( T \) as a function of its time in the oven, \( t \), in hours.
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1. The initial temperature of the roast is \( {40}^{ \circ }\mathrm{F} \), so \( {T}_{0} = {40} \) . The environment in which we are placing the roast is the \( {350}^{ \circ }\mathrm{F} \) oven, so \( {T}_{a} = {350} \) . Newton’s Law of Warming tells us \( T\left( t\right) = {350} + \left( {{40} - {350}}\right) {e}^{-{kt}} \), or \( T\left( t\right) = {350} - {310}{e}^{-{kt}} \) . To determine \( k \), we use the fact that after 2 hours, the roast is \( {125}^{ \circ }\mathrm{F} \), which means \( T\left( 2\right) = {125} \) . This gives rise to the equation \( {350} - {310}{e}^{-{2k}} = {125} \) which yields \( k = - \frac{1}{2}\ln \left( \frac{45}{62}\right) \approx {0.1602} \) . The temperature function is\n\n\[ T\left( t\right) = {350} - {310}{e}^{\frac{t}{2}\ln \left( \frac{45}{62}\right) } \approx {350} - {310}{e}^{-{0.1602t}}. \]
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Yes
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The number of people \( N \), in hundreds, at a local community college who have heard the rumor 'Carl is afraid of Virginia Woolf' can be modeled using the logistic equation\n\n\[ N\left( t\right) = \frac{84}{1 + {2799}{e}^{-t}} \]\n\nwhere \( t \geq 0 \) is the number of days after April 1,2009.\n\n1. Find and interpret \( N\left( 0\right) \).\n2. Find and interpret the end behavior of \( N\left( t\right) \).\n3. How long until 4200 people have heard the rumor?\n4. Check your answers to 2 and 3 using your calculator.
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## Solution.\n\n1. We find \( N\left( 0\right) = \frac{84}{1 + {2799}{e}^{0}} = \frac{84}{2800} = \frac{3}{100} \). Since \( N\left( t\right) \) measures the number of people who have heard the rumor in hundreds, \( N\left( 0\right) \) corresponds to 3 people. Since \( t = 0 \) corresponds to April 1,2009, we may conclude that on that day,3 people have heard the rumor.\n\n2. We could simply note that \( N\left( t\right) \) is written in the form of Equation 6.7, and identify \( L = {84} \). However, to see why the answer is 84, we proceed analytically. Since the domain of \( N \) is restricted to \( t \geq 0 \), the only end behavior of significance is \( t \rightarrow \infty \). As we’ve seen before, as \( t \rightarrow \infty \), we have \( {1997}{e}^{-t} \rightarrow {0}^{ + } \) and so \( N\left( t\right) \approx \frac{84}{1 + \text{very small }\left( +\right) } \approx {84} \). Hence, as \( t \rightarrow \infty \), \( N\left( t\right) \rightarrow {84} \). This means that as time goes by, the number of people who will have heard the rumor approaches 8400.\n\n3. To find how long it takes until 4200 people have heard the rumor, we set \( N\left( t\right) = {42} \). Solving \( \frac{84}{1 + {2799}{e}^{-t}} = {42} \) gives \( t = \ln \left( {2799}\right) \approx {7.937} \). It takes around 8 days until 4200 people have heard the rumor.\n\n4. We graph \( y = N\left( x\right) \) using the calculator and see that the line \( y = {84} \) is the horizontal asymptote of the graph, confirming our answer to part 2 , and the graph intersects the line \( y = {42} \) at \( x = \ln \left( {2799}\right) \approx {7.937} \), which confirms our answer to part 3.
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Yes
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1. If a 7 character case-sensitive \( {}^{16} \) password is comprised of letters and numbers only, find the associated information entropy.
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1. There are 26 letters in the alphabet, 52 if upper and lower case letters are counted as different. There are 10 digits (0 through 9) for a total of \( N = {62} \) symbols. Since the password is to be 7 characters long, \( L = 7 \) . Thus, \( H = 7{\log }_{2}\left( {62}\right) = \frac{7\ln \left( {62}\right) }{\ln \left( 2\right) } \approx {41.68} \).
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Yes
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Find the partial pressure of carbon dioxide in arterial blood if the \( \mathrm{{pH}} \) is 7.4.
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Solution. We set \( \mathrm{{pH}} = {7.4} \) and get \( {7.4} = {6.1} + \log \left( \frac{800}{x}\right) \), or \( \log \left( \frac{800}{x}\right) = {1.3} \) . Solving, we find \( x = \frac{800}{{10}^{1 \cdot 3}} \approx {40.09} \) . Hence, the partial pressure of carbon dioxide in the blood is about 40 torr.
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Yes
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Write the standard equation of the circle with center \( \left( {-2,3}\right) \) and radius 5.
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Here, \( \left( {h, k}\right) = \left( {-2,3}\right) \) and \( r = 5 \), so we get\n\n\[ \n{\left( x - \left( -2\right) \right) }^{2} + {\left( y - 3\right) }^{2} = {\left( 5\right) }^{2} \n\]\n\n\[ \n{\left( x + 2\right) }^{2} + {\left( y - 3\right) }^{2} = {25} \n\]
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Yes
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Graph \( {\left( x + 2\right) }^{2} + {\left( y - 1\right) }^{2} = 4 \) . Find the center and radius.
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Solution. From the standard form of a circle, Equation 7.1, we have that \( x + 2 \) is \( x - h \), so \( h = - 2 \) and \( y - 1 \) is \( y - k \) so \( k = 1 \) . This tells us that our center is \( \left( {-2,1}\right) \) . Furthermore, \( {r}^{2} = 4 \), so \( r = 2 \) . Thus we have a circle centered at \( \left( {-2,1}\right) \) with a radius of 2 . Graphing gives us
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Yes
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Complete the square to find the center and radius of \( 3{x}^{2} - {6x} + 3{y}^{2} + {4y} - 4 = 0 \) .
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\[ 3{x}^{2} - {6x} + 3{y}^{2} + {4y} - 4 = 0 \] \[ 3{x}^{2} - {6x} + 3{y}^{2} + {4y} = 4 \] add 4 to both sides \[ 3\left( {{x}^{2} - {2x}}\right) + 3\left( {{y}^{2} + \frac{4}{3}y}\right) = 4 \] factor out leading coefficients \[ 3\left( {{x}^{2} - {2x} + \underline{1}}\right) + 3\left( {{y}^{2} + \frac{4}{3}y + \underline{\underline{\frac{4}{9}}}}\right) = 4 + 3\underline{\left( 1\right) } + 3\underline{\left( \frac{4}{9}\right) } \] complete the square in \( x, y \) \[ 3{\left( x - 1\right) }^{2} + 3{\left( y + \frac{2}{3}\right) }^{2} = \frac{25}{3} \] factor \[ {\left( x - 1\right) }^{2} + {\left( y + \frac{2}{3}\right) }^{2} = \frac{25}{9} \] divide both sides by 3 From Equation 7.1, we identify \( x - 1 \) as \( x - h \), so \( h = 1 \), and \( y + \frac{2}{3} \) as \( y - k \), so \( k = - \frac{2}{3} \). Hence, the center is \( \left( {h, k}\right) = \left( {1, - \frac{2}{3}}\right) \). Furthermore, we see that \( {r}^{2} = \frac{25}{9} \) so the radius is \( r = \frac{5}{3} \).
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Yes
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Write the standard equation of the circle which has \( \left( {-1,3}\right) \) and \( \left( {2,4}\right) \) as the endpoints of a diameter.
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Solution. We recall that a diameter of a circle is a line segment containing the center and two points on the circle. Plotting the given data yields\n\n\n\nSince the given points are endpoints of a diameter, we know their midpoint \( \left( {h, k}\right) \) is the center of the circle. Equation 1.2 gives us\n\n\[ \left( {h, k}\right) = \left( {\frac{{x}_{1} + {x}_{2}}{2},\frac{{y}_{1} + {y}_{2}}{2}}\right) \]\n\n\[ = \left( {\frac{-1 + 2}{2},\frac{3 + 4}{2}}\right) \]\n\n\[ = \left( {\frac{1}{2},\frac{7}{2}}\right) \]\n\nThe diameter of the circle is the distance between the given points, so we know that half of the distance is the radius. Thus,\n\n\[ r = \frac{1}{2}\sqrt{{\left( {x}_{2} - {x}_{1}\right) }^{2} + {\left( {y}_{2} - {y}_{1}\right) }^{2}} \]\n\n\[ = \frac{1}{2}\sqrt{{\left( 2 - \left( -1\right) \right) }^{2} + {\left( 4 - 3\right) }^{2}} \]\n\n\[ = \frac{1}{2}\sqrt{{3}^{2} + {1}^{2}} \]\n\n\[ = \frac{\sqrt{10}}{2} \]\n\nFinally, since \( {\left( \frac{\sqrt{10}}{2}\right) }^{2} = \frac{10}{4} \), our answer becomes \( {\left( x - \frac{1}{2}\right) }^{2} + {\left( y - \frac{7}{2}\right) }^{2} = \frac{10}{4} \)
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Yes
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Find the points on the unit circle with \( y \) -coordinate \( \frac{\sqrt{3}}{2} \) .
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Solution. We replace \( y \) with \( \frac{\sqrt{3}}{2} \) in the equation \( {x}^{2} + {y}^{2} = 1 \) to get\n\n\[ \n{x}^{2} + {y}^{2} = 1 \n\] \n\n\[ \n{x}^{2} + {\left( \frac{\sqrt{3}}{2}\right) }^{2} = 1 \n\] \n\n\[ \n\frac{3}{4} + {x}^{2} = 1 \n\] \n\n\[ \n{x}^{2} = \frac{1}{4} \n\] \n\n\[ \nx = \pm \sqrt{\frac{1}{4}} \n\] \n\n\[ \nx = \pm \frac{1}{2} \n\] \n\nOur final answers are \( \left( {\frac{1}{2},\frac{\sqrt{3}}{2}}\right) \) and \( \left( {-\frac{1}{2},\frac{\sqrt{3}}{2}}\right) \) .
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Yes
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Graph \( {\left( x + 1\right) }^{2} = - 8\left( {y - 3}\right) \) . Find the vertex, focus, and directrix.
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Solution. We recognize this as the form given in Equation 7.2. Here, \( x - h \) is \( x + 1 \) so \( h = - 1 \) , and \( y - k \) is \( y - 3 \) so \( k = 3 \) . Hence, the vertex is \( \left( {-1,3}\right) \) . We also see that \( {4p} = - 8 \) so \( p = - 2 \) . Since \( p < 0 \), the focus will be below the vertex and the parabola will open downwards.\n\nThe distance from the vertex to the focus is \( \left| p\right| = 2 \), which means the focus is 2 units below the vertex. From \( \left( {-1,3}\right) \), we move down 2 units and find the focus at \( \left( {-1,1}\right) \) . The directrix, then, is 2 units above the vertex, so it is the line \( y = 5 \) .
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Yes
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Find the standard form of the parabola with focus \( \\left( {2,1}\\right) \) and directrix \( y = - 4 \) .
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Solution. Sketching the data yields,\n\n\n\n---\n\n\\({}^{2}\\) No, I’m not making this up.\n\n\\({}^{3}\\) Consider this an exercise to show what follows.\n\n---\n\n\n\nFrom the diagram, we see the parabola opens upwards. (Take a moment to think about it if you don’t see that immediately.) Hence, the vertex lies below the focus and has an \\( x \\) -coordinate of 2 . To find the \\( y \\) -coordinate, we note that the distance from the focus to the directrix is \\( 1 - \\left( {-4}\\right) = 5 \\) , which means the vertex lies \\( \\frac{5}{2} \\) units (halfway) below the focus. Starting at \\( \\left( {2,1}\\right) \\) and moving down \\( 5/2 \\) units leaves us at \\( \\left( {2, - 3/2}\\right) \\), which is our vertex. Since the parabola opens upwards, we know \\( p \\) is positive. Thus \\( p = 5/2 \\) . Plugging all of this data into Equation 7.2 give us\n\n\\[\n{\\left( x - 2\\right) }^{2} = 4\\left( \\frac{5}{2}\\right) \\left( {y - \\left( {-\\frac{3}{2}}\\right) }\\right)\n\\]\n\n\\[\n{\\left( x - 2\\right) }^{2} = {10}\\left( {y + \\frac{3}{2}}\\right)\n\\]
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Yes
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Graph \( {\left( y - 2\right) }^{2} = {12}\left( {x + 1}\right) \) . Find the vertex, focus, and directrix.
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Solution. We recognize this as the form given in Equation 7.3. Here, \( x - h \) is \( x + 1 \) so \( h = - 1 \) , and \( y - k \) is \( y - 2 \) so \( k = 2 \) . Hence, the vertex is \( \left( {-1,2}\right) \) . We also see that \( {4p} = {12} \) so \( p = 3 \) . Since \( p > 0 \), the focus will be the right of the vertex and the parabola will open to the right. The distance from the vertex to the focus is \( \left| p\right| = 3 \), which means the focus is 3 units to the right. If we start at \( \left( {-1,2}\right) \) and move right 3 units, we arrive at the focus \( \left( {2,2}\right) \) . The directrix, then, is 3 units to the left of the vertex and if we move left 3 units from \( \left( {-1,2}\right) \), we’d be on the vertical line \( x = - 4 \) . Since the focal diameter is \( \left| {4p}\right| = {12} \), the parabola is 12 units wide at the focus, and thus there are points 6 units above and below the focus on the parabola.
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Yes
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