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: 15

Answer

$$ f(x)=x^{4}-18x^{2}+1; \quad\quad[-4,4] $$ Critical numbers are: $0$, $-3$ and $3$. The absolute maximum, $1 $, occurs when $x=0 $ and the absolute minimum, $-80$ occurs when $x=-3$ and $ 3.$
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Work Step by Step

$$ f(x)=x^{4}-18x^{2}+1; \quad\quad[-4,4] $$ First look for critical numbers in the interval $(-4,4)$. find the points where the derivative is $0$ $$ \begin{aligned} f^{\prime}(x) &=(4)x^{3}-18(2)x+(0);\\ &=4x^{3}-36x \end{aligned} $$ Solve the equation $f^{\prime}(x)=0 $ to get $$ \begin{aligned} f^{\prime}(x)& =4x^{3}-36x=0\\ &= 4x(x^{2}-9)=0\\ \Rightarrow\quad\quad\quad\quad\quad\\ &= 4x(x-3)(x+3)=0\\ \Rightarrow\quad\quad\quad\quad\quad\\ & x =0 \quad \text {or} \quad (x-3) =0 \quad \text {or} \quad (x+3) =0 \\ \Rightarrow\quad\quad\quad\quad\quad\\ & x =0 \quad \text {or} \quad x =3 \quad \text {or} \quad x=-3 \end{aligned} $$ So, critical numbers are: $0$, $-3$ and $3$. Now, evaluate the function at the critical numbers and the endpoints: $$ \text{when } x=-4: f(-4)=(-4)^{4}-18(-4)^{2}+1 =-31 \\ \text{when } x=-3: f(-3)=(-3)^{4}-18(-3)^{2}+1 =-80 \\ \text{when } x=0: f(0)=(0)^{4}-18(0)^{2}+1 =1 \\ \text{when } x=3: f(3)=(3)^{4}-18(3)^{2}+1 =-80 \\ \text{when } x=4: f(4)=(4)^{4}-18(4)^{2}+1 =-31 \\ $$ We conclude that, the absolute maximum, $1 $, occurs when $x=0 $ and the absolute minimum, $-80$ occurs when $x=-3, 3.$
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