# Let I be any inte

It is given that  Now, f’(x) =0 The points x =1 and x = -1 divide the real line in three disjoint intervals

(-∞,-1),(-1,1) and (1,∞)

Now in interval, (-1,1)

it is clear that -1 < x < 1

x2 < 1  Therefore, f’(x) = 1- < 0 (-1,1) ~ {0}

Therefore, f is strictly decreasing on (-1,1) ~ {0}

x < -1 or 1 < x

x2 > 1  Therefore, f’(x) = 1- > 0 (-∞, -1) and (1,∞)

Therefore, f is strictly increasing in interval I disjoint from (-1,1)

Hence Proved.

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