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: 13

Answer

$$ f(x)=\frac{1}{3}x^{3}+\frac{3}{2}x^{2}-4x+1; \quad\quad[-5,2] $$ Critical numbers are: $-4$ and $1$. The absolute maximum, $\frac{59}{3} \approx 19.666 $, occurs when $x=-4 $, and the absolute minimum, $-\frac{7}{6} \approx -1.1666 $ occurs when $x=1.$
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Work Step by Step

$$ f(x)=\frac{1}{3}x^{3}+\frac{3}{2}x^{2}-4x+1; \quad\quad[-5,2] $$ First look for critical numbers in the interval $(-5,2)$. find the points where the derivative is $0$ $$ \begin{aligned} f^{\prime}(x) &=\frac{1}{3}(3)x^{2}+\frac{3}{2} (2)x-4(1)+(0);\\ &=x^{2}+3x-4 \end{aligned} $$ Solve the equation $f^{\prime}(x)=0 $ to get $$ \begin{aligned} f^{\prime}(x)& =x^{2}+3x-4=0\\ &= (x+4)(x-1)=0\\ \Rightarrow\quad\quad\quad\quad\quad\\ & (x+4) =0 \quad \text {or} \quad (x-1) =0 \\ \\ \Rightarrow\quad\quad\quad\quad\quad\\ & x =-4 \quad\text {or} \quad x =1 \\ \end{aligned} $$ So, critical numbers are: $-4$ and $1$. Now, evaluate the function at the critical numbers and the endpoints: $$ \text{when } x=-5: f(-5)=\frac{1}{3}(-5)^{3}+\frac{3}{2}(-5)^{2}-4(-5)+1\\ =\frac{101}{6} \approx 16.83333 \\ \text{when } x=-4: f(-4)=\frac{1}{3}(-4)^{3}+\frac{3}{2}(-4)^{2}-4(-4)+1\\ =\frac{59}{3} \approx 19.666 \\ \text{when } x=1: f(1)=\frac{1}{3}(1)^{3}+\frac{3}{2}(1)^{2}-4(1)+1\\ =-\frac{7}{6} \approx -1.1666 \\ \text{when } x=2: f(2)=\frac{1}{3}(2)^{3}+\frac{3}{2}(2)^{2}-4(2)+1\\ =\frac{5}{3} \approx 1.6666 \\ $$ We conclude that, the absolute maximum, $\frac{59}{3} \approx 19.666 $, occurs when $x=-4 $ and the absolute minimum, $-\frac{7}{6} \approx -1.1666 $ occurs when $x=1.$
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