Let $f$ be twice differential function and that $2f(0) = f(1)+f(-1)$, prove that there exists $c \in (-1,1)$ such that $f''(c)=0$
My attempt : Using Taylor series we get that :
$f(1) = f(0) +\frac{f'(0)}{1!} (1-0)^1 +\frac{f''(a)}{2!} (1-0)^2$
$f(-1) = f(0)+\frac{f'(0)}{1!} (-1-0)^1 + \frac{f''(b)}{2!}(-1-0)^2$
Adding together gives $f(1)+f(-1) = 2f(0) +\frac{f''(a)+f''(b)}{2}$
Which is $0 = \frac{f''(a)+f''(b)}{2}$ leading to $f''(a)=-f''(b)$ , now if $f''(a)=0$ then let $c=a$, otherwise WLG $f''(a) < 0 < f''(b)$ so by Darboux theorem the derivative must have the intermediate value property so there is $c$ such that $f''(c)=0$ , first is this proof correct ?
secondly , is there a simpler proof ?
Thanks
Your proof looks OK to me.
Attempt.
MVT:
$f(1)-f(0)+f(-1)-f(0)=0;$
$\dfrac{f(1)-f(0)}{1-0} - \dfrac{f(-1) -f(0)}{-1-0}=0;$
$f'(b)-f'(a)=0$, where
$b \in (0,1), a \in (-1,0).$
With $a <b:$
$\dfrac{f'(b)-f'(a)}{b-a}= f''(c)=0$, $c \in (a,b)$