Finding critical numbers

Differentiate to get $f'(x) = 3x^2 - 6x - 9$. This is a polynomial, so it is defined for all $x$, meaning the only critical numbers come from setting $f'(x) = 0$.

Solve $3x^2 - 6x - 9 = 0$, which simplifies to $x^2 - 2x - 3 = 0$, factoring as $(x-3)(x+1) = 0$. This gives $x = 3$ and $x = -1$, both of which lie in $(-2, 5)$.

Evaluate $f$ at the candidates $x = -2, -1, 3, 5$: $f(-2) = -8-12+18+5=3$, $f(-1) = -1-3+9+5=10$, $f(3) = 27-27-27+5=-22$, and $f(5)=125-75-45+5=10$.

Comparing the values $3, 10, -22, 10$, the largest is $10$ (attained at both $x=-1$ and $x=5$) and the smallest is $-22$ (at $x=3$). So the absolute maximum value is $10$ and the absolute minimum value is $-22$.

Since $f'(x) = 3(x-3)(x+1)$, test a point slightly less than $-1$, say $x=-1.5$: $f'(-1.5) = 3(-4.5)(-0.5) > 0$, so $f$ is increasing just before $x=-1$.

Test a point slightly greater than $-1$, say $x=0$: $f'(0) = 3(-3)(1) = -9 < 0$, so $f$ is decreasing just after $x=-1$. Since $f'$ changes from positive to negative at $x=-1$, $f$ has a relative maximum there.