# Mean Value theorem

1. Jul 18, 2010

### SpY]

1. The problem statement, all variables and given/known data
Let a>b be Real numbers and
f, g: [a,b] --> R be continuous and differentiable on (a,b)

Show g is injective on [a,b] if g'(x) != 0 for all x in (a,b)

2. Relevant equations
Rolle's theorem: Continuity and differentiability (in the conditions above) imply that
f(a) = f(b) and there exists c in (a,b) such that f'(c) = 0

3. The attempt at a solution
Well first I don't know exactly what injective means (what is "distinctness"). What I do understand is Rolle's theorem: that there is a turning point or point of zero gradient between any two points that have the same y-value (if that's right). So in this question there is no turning point or zero gradient in the interval [a,b] - but I don't know what the function is restricted to look like. I'm thinking it could be a horizontal straight line, a parabola, or a squiggly thing that starts and ends between two horizontal points. I'm really quite clueless how to prove something for all situations

If you could just give me a starting point or outline,
Thanks

2. Jul 18, 2010

### Office_Shredder

Staff Emeritus
You may have heard the term 1-1 before.

The definition for injective (1-1 is the same thing) is: if g(x)=g(y) then x=y. For example, the function y=x2 is not injective because (-1)2=12, but the function y=x3 is injective because if a3=b3 this forces a=b

In fact in your post you almost stated a solution to the problem: a re-wording of this definition is that g(x) is not injective if there are two different values of x for which g(x) is the same