Answer
$$
f(x)=x^{3}-6x^{2}+9x-8; \quad\quad[0,5]
$$
First look for critical numbers in the interval $(0,5)$.
find the points where the derivative is $0$
$$
\begin{aligned}
f^{\prime}(x) &=(3)x^{2}-6(2)x+9(1)-(0)\\
&=3x^{2}-12x+9
\end{aligned}
$$
Solve the equation $f^{\prime}(x)=0 $ to get
$$
\begin{aligned}
f^{\prime}(x)& =3x^{2}-12x+9=0\\
&= (3x-3)(x-3)=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
& (3x-3) =0 \quad \text {or} \quad (x-3) =0 \\
\\
\Rightarrow\quad\quad\quad\quad\quad\\
& 3x =3 \quad\text {or} \quad x =3 \\
\Rightarrow\quad\quad\quad\quad\quad\\
& x =1 \quad \text {or} \quad x =3 .\\
\end{aligned}
$$
So, critical numbers are: $1$ and $3$.
Now, evaluate the function at the critical numbers and the endpoints:
$$
\text{when } x=0: f(0)=(0)^{3}-6(0)^{2}+9(0)-8=-8 \\
\text{when } x=1: f(1)=(1)^{3}-6(1)^{2}+9(1)-8=-4 \\
\text{when } x=3: f(3)=(3)^{3}-6(3)^{2}+9(3)-8=-8 \\
\text{when } x=5: f(5)=(5)^{3}-6(5)^{2}+9(5)-8=12 \\
$$
We conclude that, the absolute maximum, $12$, occurs when $x=5 $ and the absolute minimum, $-8 $ occurs when $x=0, 3. $
Work Step by Step
$$
f(x)=x^{3}-6x^{2}+9x-8; \quad\quad[0,5]
$$
First look for critical numbers in the interval $(0,5)$.
find the points where the derivative is $0$
$$
\begin{aligned}
f^{\prime}(x) &=(3)x^{2}-6(2)x+9(1)-(0)\\
&=3x^{2}-12x+9
\end{aligned}
$$
Solve the equation $f^{\prime}(x)=0 $ to get
$$
\begin{aligned}
f^{\prime}(x)& =3x^{2}-12x+9=0\\
&= (3x-3)(x-3)=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
& (3x-3) =0 \quad \text {or} \quad (x-3) =0 \\
\\
\Rightarrow\quad\quad\quad\quad\quad\\
& 3x =3 \quad\text {or} \quad x =3 \\
\Rightarrow\quad\quad\quad\quad\quad\\
& x =1 \quad \text {or} \quad x =3 .\\
\end{aligned}
$$
So, critical numbers are: $1$ and $3$.
Now, evaluate the function at the critical numbers and the endpoints:
$$
\text{when } x=0: f(0)=(0)^{3}-6(0)^{2}+9(0)-8=-8 \\
\text{when } x=1: f(1)=(1)^{3}-6(1)^{2}+9(1)-8=-4 \\
\text{when } x=3: f(3)=(3)^{3}-6(3)^{2}+9(3)-8=-8 \\
\text{when } x=5: f(5)=(5)^{3}-6(5)^{2}+9(5)-8=12 \\
$$
We conclude that, the absolute maximum, $12$, occurs when $x=5 $ and the absolute minimum, $-8 $ occurs when $x=0, 3. $
