AP Calculus AB and BC: Chapter 3 – Applications of Differentiation :3.5 -The Second Derivative Test Study Notes

Solution
(a) Use the first derivative test to find the relative extreme values.
$
\begin{aligned}
& f^{\prime}(x)=15 x^4-15 x^2=15 x^2(x+1)(x-1) \\
& f^{\prime}(x)=0 \Rightarrow x=-1,0,1
\end{aligned}
$

Since $f^{\prime}$ changes from positive to negative at $x=-1$, $f(-1)=5$ is a relative maximum.
The sign of $f^{\prime}$ does not change at $x=0$, so there is no maximum or minimum.
Since $f^{\prime}$ changes from negative to positive at $x=1$, $f(1)=1$ is a relative minimum.
(b) $f$ is increasing on $(\infty,-1)$ and $(1, \infty)$. $f$ is decreasing on $(-1,1)$.
(c) $f^{\prime \prime}(x)=60 x^3-30 x=30 x\left(2 x^2-1\right)$
$
f^{\prime \prime}(x)=0 \Rightarrow x=-\frac{\sqrt{2}}{2}, 0, \frac{\sqrt{2}}{2}
$

$f^{\prime \prime}$ changes from negative to positive at $x=-\sqrt{2} / 2$, from positive to negative at $x=0$, and from negative to positive at $x=\sqrt{2} / 2$, so, $-\sqrt{2} / 2,0$, and $\sqrt{2} / 2$ are the $x$-coordinate of the points of inflection.
(d) The graph of $f$ is concave downward on $(-\infty,-\sqrt{2} / 2)$ and $(0, \sqrt{2} / 2)$.

The graph of $f$ is concave upward on $(-\sqrt{2} / 2,0)$ and $(\sqrt{2} / 2, \infty)$.