Answer
$f$ is
decreasing on $(-\infty, -1)$ and $(0,1)$ , and
increasing on $(-1,0)$ and $(1, \infty)$ .
Work Step by Step
From the graph, $f$ is
decreasing on $(-\infty, -1)$ and $(0,1)$ , and
increasing on $(-1,0)$ and $(1, \infty)$ .
Analytically,
$f(x)=x^{4}-2x^{2},$ defined everywhere,
$f^{\prime}(x)=4x^{3}-4x$, differentiable everywhere
$f^{\prime}(x)=0$
$4x^{3}-4x=0$
$4x(x-1)(x+1)=0$ .
Critical numbers: $x=0, \pm 1.$
$\left[\begin{array}{lllll}
Interval & (-\infty,-1) & (-1,0) & (0,1) & (1,\infty)\\
\text{test point} & -2 & -0.5 & 0.5 & 2\\
f^{\prime}(\text{test point}) & 4(-8)-4(-2) & 4(-1/8)-4(-1/2) & 4(1/8)-4(1/2) & 4(8)-2(2)\\
\text{sign} & - & + & - & +\\
& \searrow & \nearrow & \searrow & \nearrow
\end{array}\right]$
