Let $f : \mathbb{R} \to \mathbb{R}$ be twice differentiable Let $f : \mathbb{R} \to \mathbb{R}$ be twice differentiable. Let
$$g(x) =f(x)−f(0)-f'(0)x-[f(1)-f(0)-f'(0)]x^2.$$
Show that there is a number $t \in (0, 1)$ where $g''(t) = 0.$
Deduce that
$$
f(1) = f(0) + f
'
(0) + (1
/2)
f
''(t).
$$
Could someone explain how to proceed? What I could use to solve this.
 A: $g’’(x)=f’’(x)-2[f(1)-f(0)-f’(0)]$
By Lagrange you have that there exists $c\in [0,1]$ such that 
$f(1)-f(0)=f’(c)$ 
And there exists $b\in [0,c]$ such that 
$f’(c)-f’(0)=cf’’(b)$
so
$g’’(x)=f’’(x)-2cf’’(b)$ 
If
$f’’(b)=0$ then you have finished because 
$g’’(b)=f’’(b)-2cf’’(b)=0-2c0=0$
By contradiction suppose that, for example, $f’’(b)>0$.
Then
$(g-f)’’(x)=-2cf’’(b)<0$
So 
$(−f(0)-f'(0)x-[f(1)-f(0)-f'(0)]x^2)’’(x)<0$
and the polynomial must be decreasing
on $\mathbb{R}$.
If $f(1)-f(0)-f’(0)\neq 0$ then your polynomial it is a parabola that is not decreasing on $\mathbb{R}$ so you must have that it is 0 but is not possible because 
$0=(-f(0)-f’(0)x)’’(x)<0$
So you have  that $g’’(b)=0$ then
$0= f’’(b)-2[f(1)-f(0)-f’(0)]$
So
$f(1)=f(0)+f’(0)+\frac{1}{2}f’’(b)$
A: We have $g(0)=g(1)=0$ so there is a $c\in(0,1)$ such that $g'(c) =0$ ie $$f'(c) - f'(0)-2c(f(1)-f(0)-f'(0))=0$$ ie $$f(1)-f(0)-f'(0)=\frac{f'(c)-f'(0)}{2c}=\frac{f''(t)}{2}$$ for some $t\in(0,c)\subset (0,1)$. Thus we have $$f(1)=f(0)+f'(0)+\frac{1}{2}f''(t)$$ as desired. 
