Answer
See explanation.
Work Step by Step
\[
\left(-1+x^{2 / 3}\right)^{2}=f(x)
\]
Differentiate
\[
f^{\prime}(x)=2\left(-1+x^{2 / 3} \right) \cdot \frac{2}{3} x^{-1 / 3}=\frac{4}{3}\left(x^{1 / 3}-x^{-1 / 3}\right)
\]
Differentiate
\[
f^{\prime \prime}(x)=\frac{4}{3}\left(\frac{1}{3} x^{-4 / 3}+\frac{1}{3} x^{-2 / 3}\right)=\frac{4 x^{-4 / 3}}{9}\left(1+x^{2 / 3}\right)
\]
(a)
$f(x)$ is increasing on (-1,0)$\cup(1, \infty)$
When $f^{\prime}$ is positive, $f$ is increasing
$(\mathbf{b})$
$f(x)$ is decreasing on $(-\infty,-1) \cup(0,1)$
When $f^{\prime}$ is negative, $f$ is decreasing
$(c)$
$f(x)$ is concave up on $(-\infty, \infty)$
When $f^{\prime \prime}$ is positive, $f$ is concave up
(d)
$f(x)$ is never concave down
When $f^{\prime \prime}$ is negative, $f$ is concave down
$(\mathbf{e})$
When concavity changes and $f$ is continuous, inflection occurs
There are no inflection points
