Find the absolute extreme values for f(x)=(x+1)2+3 on the interval [-2,2].

Step 1
To find : the absolute extreme values for ${\displaystyle f\left(x\right)={(x+1)}^2+3}$ on the interval [-2, 2].
Step 2
${\displaystyle f\left(x\right)={(x+1)}^2+3}$
f′(x)=2(x+1)
To find: the critical values
Put f′(x)=0
${\displaystyle \Rightarrow 2(x+1)=0}$
${\displaystyle \Rightarrow 2x+2=0}$
${\displaystyle \Rightarrow 2x=-2}$
${\displaystyle \Rightarrow x=-1}$
Step 3
To check f(x) at the endpoints and at the critical points.
${\displaystyle f(-2)={((-2)+1)}^2+3=1+3=4}$
${\displaystyle f\left(2\right)={(2+1)}^2+3=9+3=12}$
${\displaystyle f(-1)={((-1)+1)}^2+3=3}$
Absolute maximum at x=2 and value=12
Absolute maximum at x=-2 and value=4