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The following exercise is again about the Mean Value Theorem :)

Let $f : [0,1] \rightarrow \mathbb{R}$ be continuous and differentiable on $(0,1)$. Assume that $$ \lim_{x\rightarrow 0^+} f'(x)= \lambda.$$

Show that $f$ is differentiable (from the right) at $0$ and that $f'(0)=\lambda$.

Hint: Mean Value Theorem.

What exactly is 'differentiable from the right'? Do I have to show that the limit as $x$ approaches $0$ from the right side exists? How can I do that?

How can I use the Mean Value Theorem to show that the derivative on $0$ equals $\lambda$?

Mean Value Theorem: If $f:[a,b] \rightarrow \mathbb{R}$ is continuous on $ [a,b] $ and differentiable on $ (a,b)$, then there exists a point $ c \in (a,b)$ where

$$ f'(c)= \frac{f(b)-f(a)}{b-a}.$$

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What you have is incorrect. Look here for a contradiction. math.stackexchange.com/questions/157675/… –  user17762 Oct 22 '12 at 20:00
    
"Do I have to show that the limit as $x$ approaches $0$ from the right side exists?" No, that is given to you in the assumption. –  Arkamis Oct 22 '12 at 20:02
    
@EdGorcenski $lim f'(x) = \lambda$, but I have to show that $f$ is differentiable from the right at 0. Can you explain what you mean? –  MSKfdaswplwq Oct 22 '12 at 20:04
    
@Marvis What exactly is incorrect? This is a homework assignment –  MSKfdaswplwq Oct 22 '12 at 20:05
    
"$f$ is differentiable from the right at $0$" means that $$\lim_{h\to0^+} \frac{f(h)-f(0)}h$$ exists. Use the mean value theorem to rewrite $f(h)-f(0)$. –  mrf Oct 22 '12 at 20:05

3 Answers 3

up vote 4 down vote accepted

Differentiable at $\alpha$ from right means $\displaystyle{\lim_{x \to \alpha^+} \frac{f(x)-f(\alpha)}{x-\alpha}}$ exist. Its value is $f'_+(\alpha)$.

For your case, from MVT, for each $x>0$ exist $0<y_x<x$, s.t. $f'(y_x)=\frac{f(x)-f(0)}{x}$. If you take $\displaystyle{\lim_{x \to 0^+}}$ you get your result.

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Showing that $f$ is differentiable from the right at $0$ means showing the existence of $$\lim_{x\to 0^+}\frac{f(x)-f(0)}{x-0}$$ (this of course generalizes to numbers other than $0$). In particular (in this case), that this limit is equal to $\lambda$.

For a rigorous (and fairly brief) proof, take any $\epsilon>0$. There is some $\delta>0$ such that $|f'(x)-\lambda|<\epsilon$ for all $0<x<\delta$. (Why?) Fix any $0<x<\delta$, and consider the expression $$\left|\frac{f(x)-f(0)}{x-0}-\lambda\right|.$$ What does MVT let us do with the difference quotient in the expression? Given our choice of $\delta$ and of $x$, how does our expression then compare to $\epsilon$? Since we chose our $0<x<\delta$ arbitrarily, what does that mean, by definition?

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Being differentiable at $0$ from the right means $\displaystyle{\lim_{x \to 0^+} \frac{f(x)-f(0)}{x-\alpha}}$ exist. Its value is $f'_+(0)$.

From the Mean Value Theorem, for each $x>0$ there exists a $y$ with $0<y<x$, such that $$f'(y)=\lambda=\frac{f(x)-f(0)}{x} = \frac{f(x)-f(0)}{x-0}$$

This implies that:$$\displaystyle{\lim_{x \to 0^+}} f'(y) = \displaystyle{\lim_{x \to 0^+}} \lambda= \displaystyle{\lim_{x \to 0^+}}\frac{f(x)-f(0)}{x-0} = f'_+(0) = \lambda$$

Is this a good proof? Thanks for your help all :-)

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1  
It's good (though you've got an $\alpha$ in one of the denominators, when you should have a $0$). Depending on your professor, though (or depending on the axioms and theorems you have to work with), you might need something a bit more rigorous. –  Cameron Buie Oct 22 '12 at 22:29

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