Answer the given attachment please
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Answer the given attachment please
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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
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!