# Thread: Mean Value Theorem Proof Exercise...

1. ## Mean Value Theorem Proof Exercise...

At least I believe you need the Mean Value Theorem:

Let f:[0,1]->R and g:[0,1]->R be differentiable with f(0)=g(0) and f'(x)>g'(x) for x in [0,1]. Prove that for x in (0,1], we have that f(x)>g(x).

My professor set h(x)=f(x)-g(x) and then said that h(x)>0, h'(x)>0 and that is literally all he said to explain this problem. This doesn't really make much sense to me. He is never formal about anything. Is that even remotely correct?

2. Originally Posted by zhupolongjoe
At least I believe you need the Mean Value Theorem: Let f:[0,1]->R and g:[0,1]->R be differentiable with f(0)=g(0) and f'(x)>g'(x) for x in [0,1]. Prove that for x in (0,1], we have that f(x)>g(x).

My professor set h(x)=f(x)-g(x) and then said that h(x)>0, h'(x)>0 and that is literally all he said to explain this problem. This doesn't really make much sense to me. He is never formal about anything. Is that even remotely correct?
It is correct.

If the derivative of a function is positive, is the function increasing?

3. Certainly, but is this a sufficient proof or is more needed?

4. $\displaystyle \begin{gathered} h(x) = f(x) - g(x) \hfill \\ h'(x) = f'(x) - g'(x) > 0\, \hfill \\ x \in (0,1]\, \Rightarrow \,h(x) > h(0) = 0 \hfill \\ \therefore \;f(x) > g(x) \hfill \\ \end{gathered}$