Answer
Absolute minima of $0.75$ at $x=\displaystyle \frac{1\pm\sqrt{3}}{2}$
Absolute maximum of $31$ at $x=3$
Work Step by Step
Graph $f(x)$ and $f'(x)$.
See image below, work done on desmos.com.
The derivative is zero for
$x\approx-0.366$
$x=0.5$
$x\approx 1.366$,
and we calculate $f(x)$ at the endpoints and critical points (also on the image)
$f'(x)=4x^{3}-6x^{2}+1$, and a rational zero is $x=\displaystyle \frac{1}{2}$,
so, a factor of $f'$ is $(x-\displaystyle \frac{1}{2}).$
With synthetic division, $\left[\begin{array}{rrrrrrr}
1/2 & | & 4 & -6 & 0 & 1\\
& & & 2 & -2 & -1\\
\hline & & 4 & -4 & -2 & 0
\end{array}\right]$
$f'(x)=(x-\displaystyle \frac{1}{2})(4x^{2}-4x-2)=(x-\frac{1}{2})\cdot 2(2x^{2}-2x-1)$
$=(2x-1)(2x^{2}-2x-1)$
$x=\displaystyle \frac{2\pm\sqrt{4+8}}{2(2)}=\frac{2(1\pm\sqrt{3})}{2(2)}=\frac{1\pm\sqrt{3}}{2}$
Absolute minima of $0.75$ at $x=\displaystyle \frac{1\pm\sqrt{3}}{2}$
Absolute maximum of $31$ at $x=3$
