Proving One-to-One Function & Solutions with Mean Value Theorem

[Swamifez]

Use the Mean Value Theorem to show that:

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.

 

[quasar987]

You got a).

For b), suppose there is one point such that f(x) = x. Then for all y, there exists a c such that

$$f'(c)=\frac{f(y)-x}{y-x}$$

You go from there..

 

[tacman]

a) To show that f is one-to-one on the interval (a,b), we need to prove that if f(x1) = f(x2), then x1 = x2. Let's assume that f(x1) = f(x2) for some x1, x2 in (a,b). This means that f(x1) - f(x2) = 0. By the mean value theorem, there exists a point c in (x1,x2) such that f'(c) = (f(x1) - f(x2)) / (x1 - x2) = 0. This contradicts the given condition that f '(x) is not equal to 0 for all x Element Symbol (a,b). Therefore, our assumption that f(x1) = f(x2) must be false, and thus f is one-to-one on the interval (a,b).

b) To show that f(x) = x has at most one solution, we need to prove that if f(x1) = x1 and f(x2) = x2 for some x1, x2, then x1 = x2. Let's assume that f(x1) = x1 and f(x2) = x2 for some x1, x2 in (a,b). This means that |f(x1) - f(x2)| = |x1 - x2| = 0. By the mean value theorem, there exists a point c in (x1,x2) such that f'(c) = (f(x1) - f(x2)) / (x1 - x2) = 1. But we are given that |f ' (x)| < C < 1 for all x, which means that f'(c) cannot equal 1. This is a contradiction, so our assumption that f(x1) = x1 and f(x2) = x2 must be false. Therefore, there can be at most one solution for f(x) = x in the interval (a,b).