How Can the Mean Value Theorem Be Applied to Find Specific Derivative Values?

[kathrynag]

Homework Statement

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

Homework Equations

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

 

[Mark44]

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.

 

[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

 

[Mark44]

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

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.