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Question 7.T.8: Let f : [a, b] → R be a continuous function which is differe...

Let f : [a, b] → \mathbb{R} be a continuous function which is differentiable on (a, b).

(i) If f^{′}(x) = 0  for all x ∈ (a, b), f is constant on [a, b].

(ii) If, for all x ∈ (a, b), f^{′}(x) ≠ 0,  f is injective on [a, b].

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(i) Take any two points x_{1}, x_{2} ∈ [a, b]. We shall prove that f(x_{1}) = f(x_{2}). Suppose x_{1} < x_{2}, and apply the mean value theorem to the interval to conclude that that there is a point c ∈ (x_{1}, x_{2}) such that

f(x_{2}) − f(x_{1}) = (x_{2} − x_{1})f^{′}(c).

Since c ∈ (x_{1}, x_{2}), f^{′}(c) = 0 and therefore f(x_{1}) = f (x_{2}).

(ii) If f is not injective, there would be two points x_{1}, x_{2} ∈ [a, b] such that x_{1} < x_{2} and f(x_{1}) = f(x_{2}). Applying the mean value theorem to [x_{1}, x_{2}] we would then conclude that there is a c ∈ (x_{1}, x_{2}) ⊆ (a, b) where

f^{′}(c) = \frac{f(x_{2}) − f(x_{1})}{x_{2} − x_{1}} = 0.

This contradicts the hypothesis that f^{′}(x) ≠0 for all x ∈ (a, b).

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