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: 14

Answer

$$ f(x)=\frac{1}{3}x^{3}-\frac{1}{2}x^{2}-6x+3; \quad\quad[-4,4] $$ Critical numbers are: $-2$ and $3$. The absolute maximum, $\frac{31}{3} \approx 10.3333 $, occurs when $x=-2, $ and the absolute minimum, $-\frac{21}{2} \approx -10.5 $ occurs when $x=3.$
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Work Step by Step

$$ f(x)=\frac{1}{3}x^{3}-\frac{1}{2}x^{2}-6x+3; \quad\quad[-4,4] $$ First look for critical numbers in the interval $(-4,4)$. find the points where the derivative is $0$ $$ \begin{aligned} f^{\prime}(x) &=\frac{1}{3}(3)x^{2}-\frac{1}{2}(2)x-6(1)+(0);\\ &=x^{2}-x-6 \end{aligned} $$ Solve the equation $f^{\prime}(x)=0 $ to get $$ \begin{aligned} f^{\prime}(x)& =x^{2}-x-6=0\\ &= (x-3)(x+2)=0\\ \Rightarrow\quad\quad\quad\quad\quad\\ & (x-3) =0 \quad \text {or} \quad (x+2) =0 \\ \\ \Rightarrow\quad\quad\quad\quad\quad\\ & x =3 \quad\text {or} \quad x =-2 \\ \end{aligned} $$ So, critical numbers are: $-2$ and $3$. Now, evaluate the function at the critical numbers and the endpoints: $$ \text{when } x=-4: f(-4)=\frac{1}{3}(-4)^{3}-\frac{1}{2}(-4)^{2}-6(-4)+3\\ =-\frac{7}{3} \approx -2.3333 \\ \text{when } x=-2: f(-2)=\frac{1}{3}(-2)^{3}-\frac{1}{2}(-2)^{2}-6(-2)+3\\ =\frac{31}{3} \approx 10.3333 \\ \text{when } x=3: f(3)=\frac{1}{3}(3)^{3}-\frac{1}{2}(3)^{2}-6(3)+3\\ =-\frac{21}{2} \approx -10.5 \\ \text{when } x=4: f(4)=\frac{1}{3}(4)^{3}-\frac{1}{2}(4)^{2}-6(4)+3\\ =-\frac{23}{3} \approx -7.6666 \\ $$ We conclude that, the absolute maximum, $\frac{31}{3} \approx 10.3333 $, occurs when $x=-2 $ and the absolute minimum, $-\frac{21}{2} \approx -10.5 $ occurs when $x=3.$
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