Answer
At $x=-1$, the function has a local maximum of $4$
At $x=\displaystyle \frac{6+2\sqrt{2}}{3},$ the function has a local minimum of $\approx-3.0792$.
Work Step by Step
When $ x\rightarrow-\infty$, then $ f(x)\rightarrow-\infty$
When $ x\rightarrow+\infty$, then $ f(x)\rightarrow+\infty$
There are no absolute extrema.
$y'=f'(x)=\left\{\begin{array}{ll}
-\frac{1}{4}(2x)-\frac{1}{2}, & x\leq 1\\
3x^{2}-6(2x)+8, & x\gt 1
\end{array}\right.=\left\{\begin{array}{ll}
-\frac{1}{2}x-\frac{1}{2}, & x\leq 1\\
3x^{2}-12x+8, & x\gt 1
\end{array}\right.$
We first check whether $f'(x)$ is defined at $x=1$ (whether x=1 is a critical point).
$\displaystyle \lim_{x\rightarrow 1^{-}}f(x)=-\frac{1}{2}(1)-\frac{1}{2}=-1$
$\displaystyle \lim_{x\rightarrow 1^{+}}f(x)=3(1)-12(1)+8=-1$
The one sided limits exists and are equal $\Rightarrow f'(x)$ is defined at $x=1.$
$ y'=0\quad$ when $\left\{\begin{array}{ll}
-\frac{1}{2}x-\frac{1}{2}=0, & x\leq 1\\
3x^{2}-12x+8=0, & x\gt 1
\end{array}\right.$
$\left\{\begin{array}{ll}
x+1=0, & x\leq 1\\
x=\frac{12\pm\sqrt{144-4(8)(3)}}{6}, & x\gt 1
\end{array}\right.$
$\left\{\begin{array}{ll}
x=-1, & x\leq 1\\
x=\frac{12\pm\sqrt{48}}{6}, & x\gt 1
\end{array}\right.$
$\left\{\begin{array}{ll}
x=-1, & x\leq 1\\
x=\frac{12\pm 4\sqrt{2}}{6}, & x\gt 1
\end{array}\right.$
$\left\{\begin{array}{ll}
x=-1, & x\leq 1\\
x=\frac{6\pm 2\sqrt{2}}{3}, & x\gt 1
\end{array}\right.$
$x=-1$ belongs to the interval $x\leq 1$.
$\displaystyle \frac{6+2\sqrt{2}}{3}\approx 3.1457$, and belongs to the interval $x\gt 1$
$\displaystyle \frac{6-2\sqrt{2}}{3}\approx 0.845$ is not from the interval (not a critical point)
$f(-1)=4$
$f(\displaystyle \frac{6+2\sqrt{2}}{3})\approx-3.0792$
So,
At $x=-1$, the function has a local maximum of $4$
At $x=\displaystyle \frac{6+2\sqrt{2}}{3},$ the function has a local minimum of $\approx-3.0792$.
