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One-One function and differentiability

Let f be real valued, differentiable on (a, b) and \(f'(x) \ne 0 \) for all \(x \in (a, b) \). Then f is 1-1.

True

Discussion:

Suppose f is not 1-1. Then there exists \(x_1 , x_2 \in (a, b) \) such that \(f(x_1 ) = f(x_2) \). Since f(x) is differentiable it must be continuous as well. Applying Rolles Theorem in the interval \((x_1 , x_2 ) \) we conclude that there exists a number c in this interval such that f'(c) = 0. But this contradicts the given conditions. Hence f must be 1-1

November 15, 2013

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