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Suppose $f’(a) = M$ where $M > 0$. Prove that there is a $\delta>0$ such that if $0<|x-a|<\delta$, then $\frac{f(x)-f(a)}{x-a} > M/2$.

I am probably making this problem harder than it is. I would appreciate a push in the right direction. The solution isn't coming to me. Thanks!

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If the derivative is $M$, then as $x \to a$ the difference quotient approaches $M$, and so must eventually be larger than $M/2$ when $x$ is near enough to $a$. –  coffeemath Nov 8 '12 at 15:41
    
Consider using LaTeX. –  glebovg Nov 8 '12 at 15:43
    
The way it is phrased now, this is not true. Consider $f(x)=0 \forall x \in \mathbb{R}$, then $f'(a)=0 \forall a \in \mathbb{R}$ but the difference quotient does not exist, so certainly it cannot be positive. –  gt6989b Nov 8 '12 at 15:46
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1 Answer

If $\lim \frac{f(x)-f(a)}{x-a}-\frac{M}{2}$ is positive (it is equal to $\frac{M}{2}$) then the function where you apllied the limit is positive in a reduced neighborhood of $a$. That is $\frac{f(x)-f(a)}{x-a}-\frac{M}{2}>0$ in a reduced neighborhood , as you wanted to prove.

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to gt6989b : the question asks to prove in a reduced neighborhood so the point a is not included. –  tota Nov 8 '12 at 16:43
    
This is a delta-epsilon proof. I believe that the problem should be set up as [(f(x)-f(a))/(x-a)] - M < M/2 and should show that if 0<x-a<delta, then f(x)-f(a)>0. –  user42864 Nov 9 '12 at 13:15
    
There is a well known result about limits that is : if the limit of a function is positive then the function is positive in a reduce eighborhood of the point the variable coverges, I used this result. –  tota Nov 9 '12 at 14:34
    
My professor hinted the following: f''(a)=M means lim(x>a) (f(x)-f(a))/(x-a)=M. (I understand this part) Now apply the definition of limit with epsilon=M/2. Try to deduce that if 0<x-a<delta, then f(x)-f(a)>0. (I don't understand how to do this part). –  user42864 Nov 10 '12 at 15:13
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