If $\,f'$ is continuous on $[a,a+h]$ and derivable on $(a,a+h)$, then prove that there exists a real number $c$ between $a$ and $a+h$ such that $$ f(a+h)=f(a)+hf'(a)+(h^2/2)f''(c).$$
Solution provided in the book starts with defining a function $\phi$ on $[a,a+h]$ as : $$\phi(x)=f(x)+(a+h-x)f'(x)+\cfrac{1}{2}(a+h-x)^2A$$
where $A$ is a constant determined by $\phi(a)=\phi(a+h)$ i.e., $f(a)+hf'(a)+(h^2/2)A=f(a+h)$
I am unable to understand the process, can someone help!!
This is the proof of the second-order Taylor expansion with the Lagrange form of the remainder.
We have $\phi(a) = \phi(a+h) = f(a+h)$ if we choose
$$A = \frac{f(a+h) - f(a) - f'(a)h}{\frac{1}{2}h^2}.$$
By Rolle's theorem there is a point $c \in (a,a+h)$ such that $\phi'(c) = 0.$
Note that
$$\phi'(x) = f'(x) - f'(x) + (a+h - x)f''(x) - (a+h - x)A.$$
Thus, $0 = \phi'(c) = (a+h-c)f''(c) - (a+h -c)A$ implying that $A = f''(c)$ which in turn implies
$$f(a+h) = f(a) + f'(a)h + \frac{1}{2}f''(c)h^2.$$