#### Answer

The absolute maxima are $f(2) = 18$ and $f(5) = 18$
The absolute minimum is $f(0) = -2$
The point $f(2) = 18$ is a local maximum.
The point $f(4) = 14$ is a local minimum.

#### Work Step by Step

$f(x) = x^3-9x^2+24x-2$
Note that this function is continuous and differentiable for all $x$.
We can find the points where $f'(x) = 0$:
$f'(x) = 3x^2-18x+24 = 0$
$x^2-6x+8 = 0$
$(x-2)(x-4) = 0$
$x = 2,4$
We can verify the values of the function when $f'(x) = 0$ and the endpoints of the interval $[0,5]$:
$f(0) = (0)^3-9(0)^2+24(0)-2 = -2$
$f(2) = (2)^3-9(2)^2+24(2)-2 = 18$
$f(4) = (4)^3-9(4)^2+24(4)-2 = 14$
$f(5) = (5)^3-9(5)^2+24(5)-2 = 18$
The absolute maxima are $f(2) = 18$ and $f(5) = 18$
The absolute minimum is $f(0) = -2$
The point $f(2) = 18$ is a local maximum.
The point $f(4) = 14$ is a local minimum.