Calculus with Applications (10th Edition)

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

Chapter 6 - Applications of the Derivative - 6.1 Absolute Extrema - 6.1 Exercises - Page 310: 28

Answer

$$ f(x)=x^{2} e^{-0.5x} ; \quad\quad[2,5] $$ The critical number in the given interval is: $4$. The absolute maximum, $2.16536 $ occurs when $x=4 $ and the absolute minimum, $1.47151$ occurs when $x=2.$

Work Step by Step

$$ f(x)=x^{2} e^{-0.5x} ; \quad\quad[2,5] $$ First look for critical numbers in the interval $(2,5)$. find the points where the derivative is $0$ $$ \begin{aligned} f^{\prime}(x) &=x^{2}\left(-\frac{1}{2} e^{-0.5 x}\right)+2 x e^{-0.5 x} \\ &=e^{-0.5 x}\left(2 x-\frac{1}{2} x^{2}\right) \end{aligned} $$ Solve the equation $f^{\prime}(x)=0 $ to get $$ \begin{aligned} f^{\prime}(x)& =e^{-0.5 x}\left(2 x-\frac{1}{2} x^{2}\right)=0\\ &=\frac{x}{2}e^{-0.5 x}\left(4 - x\right)=0 \Rightarrow\quad\quad\quad\quad\quad\\ & x=0 \quad \text {or} \quad(4-x)=0 \\ \Rightarrow\quad\quad\quad\quad\quad\\ & x=0 \quad \text {or} \quad x=4 \\ \end{aligned} $$ We know that $x=4 \in [2,5]$ but $0 \notin [2,5]$ So, the critical number in the given interval is: $4$. Now, evaluate the function at the critical numbers and the endpoints: $$ \begin{aligned} \text{when } x=2: \\ f(2) &=(2)^{2} e^{-0.5(2)} \\ &=\frac{4}{e} \\ &\approx 1.47151 \end{aligned} $$ $$ \begin{aligned} \text{when } x=4: \\ f(4) &=(4)^{2} e^{-0.5(4)} \\ &=\frac{16}{e^2} \\ &\approx 2.16536 \end{aligned} $$ and $$ \begin{aligned} \text{when } x=5: \\ f(5) &=(5)^{2} e^{-0.5(5)} \\ &=5^2\cdot \frac{1}{e^{2.5}}\\ &\approx 2.05212 \end{aligned} $$ We conclude that, the absolute maximum, $2.16536 $ occurs when $x=4 $ and the absolute minimum, $1.47151$ occurs when $x=2.$
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