# How to prove it using Mean Value Theorem

For $a, b \in \mathbb{R}$ with $a< b$, let $f:[a,b]\rightarrow\mathbb{R}$ be continuous on $[a,b]$ and twice differentiable on $(a,b)$. Further, assume that the graph of $f$ intersects the straight line segment joining the points $(a,f(a))$ and $(b,f(b))$ at a point $(c,f(c))$ for $a<c<b$. Show that there exists a real number $ξ\in (a,b)$ such that $f''(ξ) = 0$.

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Use Mean Value Theorem twice to get $d,e$ such that $a < d < c < e < b$ such that $\displaystyle f'(d) = \frac{f(c)-f(a)}{c-a} = \frac{f(b)-f(c)}{b-c}= f'(e)$.
Then use MVT again to get some $\xi$ such that $d < \xi < e$ with $\displaystyle f''(\xi) = \frac{f'(d) - f'(e)}{d-e} = 0$.