Question

Find the absolute maximum and minimum values of function given by $f\left(x\right)=x^2-4x+8$ in the interval $[1,5]$

Answer 1

$f\left(x\right)=x^2-4x+8$

$\therefore f^{\prime }\left(x\right)=2x-4$

$\therefore f^{\prime }\left(x\right)<0\forall x\in [1,2)$ and $f^{\prime }\left(x\right)>0\forall x\in (2,5]$

The value of $f\left(x\right)$ therefore decreases as $x$ goes from 1 to 2 and increases as $x$ goes from 2 to 5

Thus, the minimum value of $f\left(x\right)$ in the given range will be at $x=2$, which is $f\left(2\right)=4-8+8=4$

For absolute maximum, we compare the values of $f\left(x\right)$ at $x=1,5$

$f\left(1\right)=1-4+8=5$

$f\left(5\right)=25-20+8=13$

Thus, the absolute maximum of $f\left(x\right)$ in the range $x\in [1,5]$ is $13$ and the absolute minimum is $4$.