f(x) = x2
⇒ f’(x) = 2x
Now, f’(x) = 0
⇒ x = 0
⇒ x = 0 is the only critical point which could possibly be the point of local maxima or local minima of f.
⇒ f’’(0) = 2, which is positive.
Then, by second derivative test,
⇒ x = 0 is point of local maxima and local minima of f at x = 0 is f(0) = 0.
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