Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 5 - Graphs and the Derivative - 5.1 Increasing and Decreasing Functions - 5.1 Exercises - Page 261: 54

Answer

The percent of concentration of a drug in the bloodstream $x$ hours after the drug is administered is given by: $$ K(x) =\frac{4 x}{3 x^{2}+27} $$ (a) The function $K(x) $ is increasing on $ (0, 3 ).$ (b) The function $K(x) $ is decreasing on $(3 , \infty).$

Work Step by Step

The percent of concentration of a drug in the bloodstream $x$ hours after the drug is administered is given by: $$ K(x) =\frac{4 x}{3 x^{2}+27} $$ To determine any critical numbers, first use the quotient rule to find the derivative, $K^{\prime}(x)$ $$ \begin{aligned} K^{\prime}(x) &=\frac{\left(3 x^{2}+27\right) 4-4 x(6 x)}{\left(3 x^{2}+27\right)^{2}} \\ &=\frac{108-12 x^{2}}{\left(3 x^{2}+27\right)^{2}} \end{aligned} $$ To find any intervals where this function is increasing or decreasing, set $K^{\prime}(x)=0$ $$ \begin{aligned} K^{\prime}(x) &= \frac{108-12 x^{2}}{\left(3 x^{2}+27\right)^{2}}=0 \\ \Rightarrow\quad\quad\quad\quad\quad\\\ & 108-12 x^{2}=0\\ & 121(9- x^{2})=0\\ \Rightarrow\quad\quad\quad\quad\quad\\\ & (3- x)(3+x)=0\\ & x=3 \\ \text { or }\\ & x=-3\\ & [ \text {refuse, since $x$ is represents time, must be at least 0}] \end{aligned} $$ so $3$ is the only critical number. Now, we can use the first derivative test. Check the sign of $K^{\prime}(x)$ in the intervals: $$ (0, 3 ), \quad (3, \infty), $$ (1) Test a number in the interval $(0, 3 ) $ say $1$: $$ \begin{aligned} K^{\prime}(1) &=\frac{108-12 (1)^{2}}{\left(3 (1)^{2}+27\right)^{2}}\\ &=\frac{8}{75} \\ &\approx 0.1066\\ & \gt 0 \end{aligned} $$ we see that $K^{\prime}(x)$ is positive in that interval, so $K(x)$ is increasing on $(0, 3)$. (2) Test a number in the interval $(3, \infty ) $ say $4$: $$ \begin{aligned} K^{\prime}(4) &=\frac{108-12 (4)^{2}}{\left(3 (4)^{2}+27\right)^{2}}\\ &=-\frac{28}{1875} \\ &\approx -0.01493\\ & \lt 0 \end{aligned} $$ we see that $K^{\prime}(x)$ is negative in that interval, so $K(x)$ is decreasing on $(3, \infty )$. So, (a) The function $K(x) $ is increasing on $ (0, 3 ).$ (b) The function $K(x) $ is decreasing on $(3 , \infty).$
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