Answer
Increasing on the interval$ (-\infty, -2)$, and $(2,\infty)$,
Decreasing on the interval $(-2, 2)$
Work Step by Step
The graph appears to be Increasing on the interval$ (-\infty, -2)$, and $(2,\infty)$,
Decreasing on the interval $(-2, 2)$
To solve analytically, set the derivative equal to zero
$y=\frac{x^3}{4}-3x$
$y'=\frac{3}{4}x^2 -3$
$0=\frac{3}{4}x^2 -3$
$4=x^2$
$x= ^+_- 2$
Use test values on the three intervals, $ (-\infty, -2)$, $(-2, 2)$, and $(2,\infty)$ to find that the function increases, decreases, then increases on the respective intervals, using the first derivative test.
