## Calculus with Applications (10th Edition)

$$f(x)=2x^{3}-4x^{2}+2$$ Critical numbers: $0, \frac{4}{3}$. By Part 2 of the second derivative test, $0$ leads to a relative maximum. Also, by Part 1 of the second derivative test, $\frac{4}{3}$ leads to a relative minimum.
$$f(x)=2x^{3}-4x^{2}+2$$ First, find the points where the derivative is $0$ Here \begin{aligned} f^{\prime}(x) &=2(3)x^{2}-4(2)x+0\\ &=6x^{2}-8x\\ & =2x(3x-4) \end{aligned} Solve the equation $f^{\prime}(x)=0$ to get \begin{aligned} f^{\prime}(x) =2x(3x-4) &=0\\ \Rightarrow\quad\quad\quad\quad\quad\quad\quad\quad\quad\\ x =0 \quad &\text {and} \quad x =\frac{4}{3} \\ \end{aligned} Critical numbers: $0, \frac{4}{3}$. Now use the second derivative test. The second derivative is $$f^{\prime\prime}(x)=6(2)x-8=12x-8.$$ Evaluate $f^{\prime\prime}(x)$ first at $0$, getting $$f^{\prime\prime}(0)=12(0)-8=-8 \lt 0.$$ so that by Part 2 of the second derivative test, $0$ leads to a relative maximum. Also, when $x=\frac{4}{3}$, $$f^{\prime\prime}(\frac{4}{3})=12(\frac{4}{3})-8=16-8=8 \gt 0.$$ so that by Part 1 of the second derivative test, $\frac{4}{3}$ leads to a relative minimum.