Answer
See the explanations.
Work Step by Step
\[
\begin{aligned}
&17-16 x^{2}+4 x^{4}= f(x) \\
&-32 x+16 x^{3}= f^{\prime}(x) \\
&=\left(-2+x^{2}\right)16 x
\end{aligned}
\]
$x=\pm \sqrt{2}, 0$ are critical points. Both the points are stationary points, as the given function is differentiable.
