Calculus with Applications (10th Edition)

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

Chapter 5 - Graphs and the Derivative - 5.3 Higher Derivatives, Concavity, and the Second Derivative Test - 5.3 Exercises - Page 284: 62

Answer

$$ f(x)=x^{3} $$ Critical number: $0$. the second derivative test gives no information about extrema, by using the first derivative test, we find that: Neither a relative maximum nor relative minimum occurs at $x=0$

Work Step by Step

$$ f(x)=x^{3} $$ First, find the points where the derivative is $0$ Here $$ \begin{aligned} f^{\prime}(x) &=3x^{2}\\ \end{aligned} $$ Solve the equation $f^{\prime}(x)=0 $ to get $$ \begin{aligned} f^{\prime}(x)& =3x^{2} =0\\ \Rightarrow\quad\quad\quad\quad\quad\\ x &=0\\ \end{aligned} $$ Critical number: $0$. Now use the second derivative test. The second derivative is $$ f^{\prime\prime}(x)=3(2)x=6x. $$ Evaluate $f^{\prime\prime}(x)$ at $0$, getting $$ f^{\prime\prime}(0)=6(0)=0 . $$ then the test gives no information about extrema, so we can use the first derivative test. Check the sign of $f^{\prime}(x)$ in the intervals $$ (-\infty, 0 ), \quad (0,\infty). $$ (1) Test a number in the interval $(-\infty, 0 )$ say $ -1$: $$ \begin{aligned} f^{\prime}(-1) =3(-1)^{2} \\ &=3(1) \\ &=3 \\ &\gt 0 \end{aligned} $$ to see that $ f^{\prime}(x)$ is positive in that interval, so $f(x)$ is increasing on $(-\infty, 0 )$ (2) Test a number in the interval $(0, \infty )$ say $1$: $$ \begin{aligned} f^{\prime}(1) =3(1)^{2} \\ &=3(1) \\ &=3 \\ &\gt 0 \end{aligned} $$ to see that $ f^{\prime}(x)$ is positive in that interval, so $f(x)$ is also increasing on $(0, \infty ).$ So that by the first derivative test, since $f(x)$ is increasing for all $x$, then neither a relative maximum nor relative minimum occurs at $x=0.$
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