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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 $a$.2012-11-08
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    Consider using [LaTeX](http://meta.math.stackexchange.com/questions/107/faq-for-math-stackexchange/117#117).2012-11-08
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    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.2012-11-08

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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.2012-11-08
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    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 00.2012-11-09
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    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.2012-11-09
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    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 00. (I don't understand how to do this part).2012-11-10