$f$ ia continuously differentiable function with $f'(c)=0$............................. Let $f$ be a continuously differentiable  function  on  $[a,b].$  There is a number $c$ in  $(a,b]$ such that  $f'(c)=0.$ Then prove that there is a fixed number $\xi\in(a,b)$ such that $$f'(\xi)={{f(\xi)-f(b)}\over {b-a}}.$$
How do I begin it , just give me some hints please .
 A: As Wojowu has pointed out, the question, as written, is false.  The counterexample is, $f(x)=x^3,a=−1,b=1$.  Then clearly, $f'(0) = 0$ for $0\in[-1, 1]$; however, $$f'(\zeta)=\frac{f(\zeta)-f(b)}{b-a}\mathrm{\ becomes\ }3\zeta^2 = \frac{\zeta^3-1}{1-^-1}$$ or $\zeta^3-6\zeta^2-1=0$.  This has no solutions in [-1, 1].
My interpretation of the question is, if you meant $$f'(\zeta)=\frac{f(b)-f(a)}{b-a}$$
First, modify $g(x) = f(x) - f(a) + \frac{x-a}{b-a}(f(a)-f(b))$.  (possibly something very similar, depending on the exact statement of the question, as discussed in the comments).
Then $g(a) = 0 = g(b)$, and you are looking for a number so that $g'(\xi) = 0$.  Since $g$ is also continuously differentiable, then there is a theorem that you can use to tell you that.
A: I believe the second statement is not true. Take the the identity, then the equation becomes $$1=\frac{\xi-b}{b-a}$$ which is impossible since because the right side is negative. 
If we change the -f(b) to a -f(a) then this is still not true, because for the identity the equation holds iff $\xi = b$
