Mean value theorem (maybe) Here is one problem in my midterm test in introduction to real analysis
Let $f$ be continuous function in $[a,b]$, $a>0$ and differentiable in $(a,b)$ show that there exist $c$ in $(a,b)$ such that $$\frac{af(b)−bf(a)}{a−b}=f(c)−cf′(c)$$
Thanks.
I'm sorry for my lack in English.
 A: Consider the function $g(x)=\frac{f(x)}{x}$ and the function $t(x)=\frac{1}{x}$.For $x>0$ Both the functions are differentiable. 
By Cauchy's mean value theorem. There exist a $c$ in $\left(a,b\right)$ such that 
$$\frac{g(b)-g(a)}{t(b)-t(a)}=\frac{g^{'}{(c)}}{t^{'}{(c)}}$$.
And this implies for non zero $c$ (as $a>0$)
$$\frac{af(b)−bf(a)}{a−b}=f(c)−cf′(c)$$
A: Something similar can be shown, making it possible the problem was not copied correctly.
Since $f$ is defined and continuous in $[a,b]$ where $a>0$, and differentiable in $(a,b)$, the function $g(x)=f(x)/x$ is also continuous and differentiable on the same intervals. So by the mean value theorem there is $c \in (a,b)$ for which 
$$\frac{g(b)-g(a)}{b-a}=g'(c).$$
Use of the quotient rule for $g'$ and algebra then gives
$$\frac{af(b)-bf(a)}{ab(a-b)}=\frac{f(c)-cf'(c)}{c^2}.$$
The difference between this and the question of the post is the extra factor of $ab$ in the denominator, and the extra copy of $c^2$ on the right. So it seems maybe the question was copied incorrectly, since looking at $g(x)=f(x)/x$ gets one so close to the statement in the problem. However maybe a slight adjustment to the definition of $g$ could cancel out the extra $ab$ and $c^2$ factors here, making the original statement hold as is. I'll try for that, but leave this up for now.
A: I have got the right answer..
Thanks for all comment and your response.
The cauchy's mean value theorem are explained in the next chapter(after the midterm test).
let define $g(x)=x{f(\frac{1}{x})}$ for $x$ in $[\frac{1}{b},\frac{1}{a}]$
by using mean value theorem, there exist $\frac{1}{c}$ in $(\frac{1}{b},\frac{1}{a})$ such that $\frac{\frac{1}{a}f(a)-\frac{1}{b}f(b)}{\frac{1}{a}-\frac{1}{b}}=f(c)-cf'(c)$
equivalent with there exist $c$ in $(a,b)$ such that $\frac{af(b)-bf(a)}{a-b}=f(c)-cf'(c)$
And we're done..
