#### Answer

$$
f(x)=x^{3}-9 x^{2}+24 x-2, \quad [0,5],
$$
The local maximum value is
$$ f(2)=18$$
and the local minimum value is
$$f(4)=14 $$
The values of $f$ at the endpoints of the interval are
$$
f(0)=-2 \text { and } f(5)=18
$$
Thus,
$$
f(0)=-2
$$
is the absolute minimum value,
$$
f(2)=f(5)=18
$$
is the absolute maximum value.

#### Work Step by Step

$$
f(x)=x^{3}-9 x^{2}+24 x-2, \quad [0,5],
$$
Since $ f$ is continuous on [0,5],we can use the Closed Interval Method
$$
f(x)=x^{3}-9 x^{2}+24 x-2
$$
$$
\begin{aligned}
f^{\prime}(x) &=3 x^{2}-18 x+24 \\
& =3\left(x^{2}-6 x+8\right) \\
&=3(x-2)(x-4)
\end{aligned}
$$
Since $f^{\prime}(x)$ exists for all $x$ the only critical numbers of $f$ occur when $f^{\prime}(x)=0$ that is,
$$
x=2 \text { or } x=4 .
$$
For $0 \lt x \lt 2 $
$$ f^{\prime}(x) \gt 0$$
For $ 2 \lt x \lt 4 $
$$f^{\prime}(x) \lt 0$$
For $ 4 \lt x \lt 5 $
$$f^{\prime}(x) \gt 0$$
so
$$f(2)=18$$
is a local maximum value and
$$f(4)=14 $$
is a local minimum value.
The values of $f$ at the endpoints of the interval are
$$
f(0)=-2 \text { and } f(5)=18
$$
Thus,
$$
f(0)=-2
$$
is the absolute minimum value,
$$
f(2)=f(5)=18
$$
is the absolute maximum value.