Answer :

f(x) = x^{2}

⇒ 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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