Use the Mean Value Theorem to show that:(adsbygoogle = window.adsbygoogle || []).push({});

a)Suppose f is a diferentiable function on the interval a < b, and suppose f '(x) is not equal to 0 for all x Element Symbol (a,b). Show that f is one-to-one on the interval (a,b).

b) Assume that |f ' (x)| < C < 1 for all x. Show that f (x) = x has at most one solution.

So far I got-> Suppose f is not one-one on the interval then there exists u, v in (a,b) such that f(u)=f(v).

Then by the mean value theorem there exists a point w in (u,v) such that f'(w)=0, a contradiction.

I don't know where to go after that.

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# MVT Proof, please help

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