1. ## Grade my derivative proof

My professor has this proof in his notes and I would appreciate some feedback. It appears in the Mean Value Theorem section but I'm not sure if I need to incorporate that or not.

Let $\displaystyle I$ be an interval, suppose $\displaystyle f$ and $\displaystyle g$ are differentiable on $\displaystyle I$. If $\displaystyle f'(x)\leq g'(x)$ on $\displaystyle (a,b)$ and $\displaystyle c$ is in $\displaystyle I$, then prove that $\displaystyle f(x)-f(c)\leq g(x)-g(c)$ for all $\displaystyle x$ in $\displaystyle I$.

My attempt:
Let $\displaystyle f'(c)=(f(x)-f(c))/(x-c)$ and $\displaystyle g'(c)=(g(x)-g(c))/(x-c)$ as $\displaystyle x\rightarrow c$. Then, $\displaystyle f'(x)\leq g'(x)$ implies that $\displaystyle (f(x)-f(c))/(x-c)\leq g(x)-g(c))/(x-c)$. This implies that $\displaystyle f(x)-f(c)\leq g(x)-g(c)$ as $\displaystyle x\rightarrow c$ as desired.

2. I think it's right! (x!=c)

3. What happens if $\displaystyle \displaystyle x - c < 0$. Surely that would then mean that $\displaystyle \displaystyle f(x) - f(c) \geq g(x) - g(c)$...

I don't think you can omit this case, because derivatives are found using limits (which by the way, you need to write in your proof), which means you need to check what happens as $\displaystyle \displaystyle x \to c$ from both sides...

4. I did not bother with thread before because I did not follow any of it.
For one thing, it is not clear what the relation $\displaystyle I~\&~(a,b)$ have to one another.
But as stated the question is not true.
Consider $\displaystyle I=(1,10)$, $\displaystyle f(x)=x^2~,~g(x)=x^3,~\&~c=4$.
Clearly $\displaystyle \left( {\forall x \in I} \right)\left[ {f'(x) \leqslant g'(x)} \right]$.
But if $\displaystyle 1<x<4$ then $\displaystyle {f(x) - f(4) > g(x) - g(4)}$.

5. So are you saying that it can't be proven at all because it is always false or do I need to approach it another way?

I was looking at it again and I was wondering if you could use the Mean Value Theorem. We know that on some interval, $\displaystyle I$, if $\displaystyle f'(x)\leq g'(x)$ then the average rate of change on that interval will be greater for $\displaystyle g$. So I guess what I am asking is what if we assumed that $\displaystyle c$ was not some random point but the point that makes $\displaystyle (g(a)-g(b))/(a-b)=g'(c)$ for some $\displaystyle c\in (a,b)$ true? Then is the statement still false?

6. Originally Posted by zebra2147
So are you saying that it can't be proven at all because it is always false or do I need to approach it another way?
What I am saying is that you need to look closely at the statement.
Originally Posted by zebra2147
Let $\displaystyle I$ be an interval, suppose $\displaystyle f$ and $\displaystyle g$ are differentiable on $\displaystyle I$. If $\displaystyle f'(x)\leq g'(x)$ on $\displaystyle (a,b)$ and $\displaystyle c$ is in $\displaystyle I$, then prove that $\displaystyle f(x)-f(c)\leq g(x)-g(c)$ for all $\displaystyle x$ in $\displaystyle I$.
Did you write in correctly?
It is not clear what the relation of $\displaystyle I$ is to $\displaystyle (a,b)$.
Is $\displaystyle c$ any point in $\displaystyle I=(a, b)~?$
If so, I gave a counter-example. So it is false.

7. The way that the question is stated then I agree it is false. But I am wondering if we do not allow $\displaystyle c$ to be ANY point but instead the only point that satisfies $\displaystyle (g(a)-g(b))/(a-b)=g'(c)$. Then is the statement still false?

8. Originally Posted by zebra2147
The way that the question is stated then I agree it is false. But I am wondering if we do not allow $\displaystyle c$ to be ANY point but instead the only point that satisfies $\displaystyle (g(a)-g(b))/(a-b)=g'(c)$. Then is the statement still false?
Try to prove it true for all $\displaystyle x\in (c,b).$

9. Alright, you were right of course... I talked to my professor and he said that I=(a,b) and that c<x. Does that change anyone's opinion of my proof?