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So let the function be defined like this: $$f:[a,b]\rightarrow\mathbb{R}, f(a)=a, f(\frac{a+b}{2})=2a, f(b)=3a$$ Also the function is double differentiable on $(a,b)$

So know i want to prove that $ \exists x_{0}$ such that $f''(x_{0})=0$

So i thought i could first use something to describe first derivative, but i somehow cannot do that, so i can't prove that for second derivative, if i don't know anything of first yet. Any help would be appreciated.

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Use the mean value theorem to show that there is at least two points, one between $a$ and $1/2(a+b)$ and one between that point and $b$ where the derivative evaluated at those points is equal. Then apply Rolle's theorem for the second derivative. Idk how into detail you want me to go since you might just want a hint...

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  • $\begingroup$ i think it's enough, although it owuld be helpful to see more for the first part $\endgroup$ – MathIsTheWayOfLife Aug 5 '16 at 7:53
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    $\begingroup$ Just apply the mean value theorem to the intervals [a,(a+b)/2] and [(a+b)/2,b] to show that there are at least two points c,d: f'(c)=f'(d) then apply Rolle's thm to f' two show that there is at least one point where f"(x)=0 $\endgroup$ – Praise Existence Aug 5 '16 at 7:58
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Hint : consider $g(x) = f(x)-\frac{2a}{b-a}(x-a)-a$. We have $g(a)=g\left(\frac{a+b}{2}\right)=g(b)=0$.

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  • $\begingroup$ thank you, it's of much help $\endgroup$ – MathIsTheWayOfLife Aug 5 '16 at 7:35

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