Question
In interval $[ 1 , 5]$ Find the absolute maximum and absolute minimum values of given function $f(x)=x^2-4 x+8$.

Answer

We know :$
\begin{aligned}
f(x) & =x^2-4 x+8 \\
\text { or } \quad f^{\prime}(x) & =2 x-4=2(x-2)
\end{aligned}
$
Put $f^{\prime}(x)=0$ we get $x=2$
We find the values of $f$ at $x=1, x=2$, and $x=5$.
now$
\begin{aligned}
\therefore f(x) & =x^2-4 x+8 \\
\therefore f(1) & =(1)^2-4 \times 1+8=5 \\
f(2) & =(2)^2-4 \times 2+8=4 \\
f(5) & =(5)^2-4 \times 5+8=13
\end{aligned}
$
Thus we find that in $[1,5]$ function at $x=5$
absolute maximum value is 13 and at $x=2$
absolute minimum value is 4 .

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