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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.

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$g’’(x)=f’’(x)-2[f(1)-f(0)-f’(0)]$

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

$f(1)=f(0)+f’(0)+\frac{1}{2}f’’(b)$

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

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