Finding critical numbers

Differentiate to get $f'(x) = 6x^2 - 30x + 24$. Factor out $6$ to get $f'(x) = 6(x^2 - 5x + 4) = 6(x-1)(x-4)$.

Setting $f'(x) = 0$ gives $x = 1$ and $x = 4$, both of which lie in $(0,6)$. Since $f'(x)$ is a polynomial, it is defined everywhere, so these are the only critical numbers.

On $(0,1)$, pick $x=0.5$: $f'(0.5) = 6(-0.5)(-3.5) > 0$, so $f$ is increasing. On $(1,4)$, pick $x=2$: $f'(2) = 6(1)(-2) < 0$, so $f$ is decreasing. On $(4,6)$, pick $x=5$: $f'(5) = 6(4)(1) > 0$, so $f$ is increasing again.

Since $f'$ changes from positive to negative at $x=1$, $f$ has a local maximum there. Since $f'$ changes from negative to positive at $x=4$, $f$ has a local minimum there.

Evaluate $f$ at the endpoints and critical numbers: $f(0) = 7$, $f(1) = 2 - 15 + 24 + 7 = 18$, $f(4) = 128 - 240 + 96 + 7 = -9$, and $f(6) = 432 - 540 + 144 + 7 = 43$.

Comparing the values $7, 18, -9, 43$, the largest is $43$ at $x=6$ and the smallest is $-9$ at $x=4$. So the absolute maximum is $43$ and the absolute minimum is $-9$.