# F'(c)=f(b)-f(a)/b-a proof

1. Dec 16, 2008

### kathrynag

1. The problem statement, all variables and given/known data
Suppose f:[0,2]-->R and g:[0,1]-->R be differentable. f(0)=0, f(1)=2, f(2)=2. Prove that there is c such that f'(c)=0, f'(c)=2, f'(c)=3/2

2. Relevant equations

3. The attempt at a solution
[f(2)-f(1)]/(2-1)=0
[(f(1)-f(0)]/(1-0)]=2
I don't know what those mean and how to show for 3/2

2. Dec 16, 2008

### Staff: Mentor

Is there some reason g is included in the problem you show? Also, you have f(1) = 2 and f(2) = 2. There's not a typo there is there?

In your attempt to solve this problem, you have [f(2) - f(1)]/(2 - 1) = 0. What does the Mean Value Theorem say about the value of f' at some point in the interval (0, 2)? (These problems are exercises in the application of the MVT.)
Same for your second problem.
Based on the information you show, I don't see a way of showing that f'(c) = 3/2, unless you have written the given information incorrectly.

3. Dec 16, 2008

### kathrynag

g shouldn't even be in there.
Both f(1) and f(2)=2
There is c in (0,2) such that f'(c)=f(b)-f(a)/b-a

4. Dec 16, 2008

### Staff: Mentor

By the MVT, there is a number c in (0, 1) such that f'(c) = [f(1) - f(0)]/(1 - 0) = 2/1 = 2
You can use the same idea to show that for another number c in (1, 2), f'(c) = 0.

If you knew that f' was continuous, you could use the Intermediate Value Theorem to show that for yet another number c, f'(c) = 3/2. The idea here is that, since the derivative is equal to 2 at some point, and equal to 0 at another point, the derivative has to take on all values between 0 and 2, provided that the derivative is continuous.