# 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.

$$g’’(x)=f’’(x)-2[f(1)-f(0)-f’(0)]$$

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

$$f(1)=f(0)+f’(0)+\frac{1}{2}f’’(b)$$

• so the last expression is equal to 0? – The Poor Jew Feb 25 at 11:01
• One moment please – Federico Fallucca Feb 25 at 11:07

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.