Prove using lagrange's mean value theorem 
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*If f' is continuous on [a,a+h] and derivable on (a,a+h) prove that  there exists  a real number  c between a and a+h such that  $f(a+h)=f(a)+hf'(a)+{\frac{h^{2}}{2}}f''(c)$.


I used lagrange's  mean  value theorem $f(x)$ will be also be continuous  in $[a,a+h]$ and differentiable in (a,a+h) hence there exists $\delta \epsilon$ $(a,a+h)$ such that 
$f'(\delta)={\frac{f(a+h)-f(a)}{h}}$
Similarly since LMVT is applicable to  $f'(x)$ and there exists  $\gamma \epsilon$ ($\delta$,a)
$f''(\gamma)={\frac{f'(\delta)-f'(a)}{\delta-a}}$ . 
But how do I prove that $f"(\gamma)= {\frac{h}{2}}f''(c)$ 
I got the above conclusion above by the following steps
${\frac{f(a+h)-f(a)}{h}}=f'(c)+{\frac{h}{2}}f''(c)$
${\frac{f'(\delta)-f'(a)}{\delta-a}}={\frac{hf''(c)}{2}}$
But I don't know how to proceed further can someone please help me with this question

It can be shown by MVT on the integral of the error.
https://brilliant.org/wiki/taylors-theorem-with-lagrange-remainder/
