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Let $ f$ be continuous on $[a,b] $ and assume the second derivative $f''$ exists on (a,b). The graph of $f$ and the line segment joining the points $(a,f(a))$ and $(b,f(b))$ intersect at a point $(x_{0},f(x_{0}))$ where $a<x_{0}<b$. Show that there exists a point $c \in (a,b) $ such that $f''(c) = 0$.

Lord_Farin
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Redsbefall
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  • Try drawing a picture. Can you see an inflection point? – Ian Coley Oct 01 '13 at 10:20
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    Hello, welcome to Math.SE. Please, try to make the title of your questions more informative. E.g., Why does $a\le b$ imply $a+c\le b+c$? is much more useful for other users than A question about inequality. For more information on choosing a good title, see this post. – Lord_Farin Oct 01 '13 at 10:21
  • Of course, thanks for the good suggestion Lord_Farin – Redsbefall Oct 01 '13 at 10:35

2 Answers2

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Hint: We have $$ \frac{f(x_0) - f(a)}{x_0 - a} = \frac{f(b) - f(x_0)}{b-x_0} $$ as the three points $(a, f(a))$, $(x_0, f(x_0))$ and $(b, f(b))$ form a line. Now apply the mean value theorem to $[a, x_0]$ and $[x_0, b]$. Rolle's theorem applied to $f'$ will do the trick.

martini
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Hint: There exist $x_1 \in (a,x_0)$, $x_2\in (x_0,b)$ with $f'(x_1)=\dfrac{f(x_0)-f(a)}{x_0-a}=\dfrac{f(b)-f(x_0)}{b-x_0}=f'(x_2).$

njguliyev
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