Question

Find the absolute maximum value and the absolute minimum value of the following function in the given interval :

$f\left(x\right)=(x-1{)}^2+3,x\in [-3,1]$

Answer 1

$f\left(x\right)=(x-1{)}^2+3$

Therefore, $f^{\prime }\left(x\right)=2(x-1)$

Now,

$f^{\prime }\left(x\right)=0⟹x=1$

Then, we evaluate the value of f at critical point $x=1$ and at the end points of the interval $[-3,1]$.

$f\left(1\right)=3$

$f(-3)=19$

Hence, the absolute maximum value of f on $[-3,1]$ is $19$ occurring at $x=-3$ and the minimum value of f on $[-3,1]$ is $3$ occurring at $x=1$.