Answer
The absolute maximum is $f(1) = ln(3)$
The absolute minimum is $f(-\frac{1}{2}) = ln(\frac{3}{4})$
Work Step by Step
$f(x) = ln~(x^2+x+1)$
We can find the points when $f'(x) = 0$:
$f(x) = ln~(x^2+x+1)$
$f'(x) = \frac{2x+1}{x^2+x+1} = 0$
$2x+1 = 0$
$x = -\frac{1}{2}$
We can find the value of $f(-\frac{1}{2})$ and the values of the function at the endpoints of the interval $[-1, 1]$:
$f(-\frac{1}{2}) =ln~[(-\frac{1}{2})^2+(-\frac{1}{2})+1] = ln(\frac{3}{4})$
$f(-1) =ln~[(-1)^2+(-1)+1] = ln(1)$
$f(1) =ln~[(1)^2+(1)+1] = ln(3)$
The absolute maximum is $f(1) = ln(3)$
The absolute minimum is $f(-\frac{1}{2}) = ln(\frac{3}{4})$
