Answer
The absolute maximum is $f(1) = \sqrt{3}$
The absolute minimum is $f(-\frac{1}{2}) = \frac{\sqrt{3}}{2}$
There is no local maximum.
$f(-\frac{1}{2}) = \frac{\sqrt{3}}{2}$ is a local minimum.
Work Step by Step
$f(x) = \sqrt{x^2+x+1}$
Note that this function is continuous on the interval $[-2,1]$ and differentiable on the interval $(-2,1)$.
We can find the points where $f'(x) = 0$:
$f'(x) = \frac{2x+1}{2\sqrt{x^2+x+1}}=0$
$2x+1 = 0$
$x = -\frac{1}{2}$
We can verify the value of the function when $f'(x) = 0$ and at the endpoints of the interval $[-2,1]$:
$f(-2) = \sqrt{(-2)^2+(-2)+1} = \sqrt{3}$
$f(-\frac{1}{2}) = \sqrt{(-\frac{1}{2})^2+(-\frac{1}{2})+1} = \frac{\sqrt{3}}{2}$
$f(1) = \sqrt{(1)^2+(1)+1} = \sqrt{3}$
The absolute maximum is $f(1) = \sqrt{3}$
The absolute minimum is $f(-\frac{1}{2}) = \frac{\sqrt{3}}{2}$
There is no local maximum.
$f(-\frac{1}{2}) = \frac{\sqrt{3}}{2}$ is a local minimum.