Question

Answer the given attachment please

Answer 1

absolute extreme of f(X)=x^2

Here the interval is a subset of like fraction and the function is a square

extreme=

Maxima=(-2)^2=(2)^2=4

minima=(0)^2=0

Answer 2

Step-by-step explanation:

Given: f(x) = x²

Then, f'(x) = 2x.

The only critical point is x = 0.

It is indeed in the interval [-2,2] so list 0 and the end points.

x      f(x)

0     0

-2     4

2      4

So (0,0) is the absolute minimum and (2,4) is the absolute maximum.

Hope it helps!