## Calculus with Applications (10th Edition)

Published by Pearson

# Chapter 5 - Graphs and the Derivative - 5.3 Higher Derivatives, Concavity, and the Second Derivative Test - 5.3 Exercises - Page 284: 59

#### Answer

$$f(x)=3x^{3}-3x^{2}+1$$ Critical numbers: $0, \frac{2}{3}$. By Part 2 of the second derivative test, $0$ leads to a relative maximum. Also, by Part 1 of the second derivative test, $\frac{2}{3}$ leads to a relative minimum.

#### Work Step by Step

$$f(x)=3x^{3}-3x^{2}+1$$ First, find the points where the derivative is $0$ Here \begin{aligned} f^{\prime}(x) &=3(3)x^{2}-3(2)x+0\\ &=9x^{2}-6x\\ & =3x(3x-2) \end{aligned} Solve the equation $f^{\prime}(x)=0$ to get \begin{aligned} f^{\prime}(x) =3x(3x-2) &=0\\ \Rightarrow\quad\quad\quad\quad\quad\quad\quad\quad\quad\\ x =0 \quad &\text {and} \quad x =\frac{2}{3} \\ \end{aligned} Critical numbers: $0, \frac{2}{3}$. Now use the second derivative test. The second derivative is $$f^{\prime\prime}(x)=9(2)x-6=18x-6.$$ Evaluate $f^{\prime\prime}(x)$ first at $0$, getting $$f^{\prime\prime}(0)=18(0)-6=-6 \lt 0.$$ so that by Part 2 of the second derivative test, $0$ leads to a relative maximum. Also, when $x=\frac{2}{3}$, $$f^{\prime\prime}(\frac{2}{3})=18(\frac{2}{3})-6=6 \gt 0.$$ so that by Part 1 of the second derivative test, $\frac{2}{3}$ leads to a relative minimum.

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.