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Suppose $f$ is a thrice differentiable function on $\mathbb {R}$ such that $f'''(x) \gt 0$ for all $x \in \mathbb {R}$. Using Taylor's theorem show that

$f(x_2)-f(x_1) \gt (x_2-x_1)f'(\frac{x_1+x_2}{2})$ for all $x_1$and $x_2$ in $\mathbb {R}$ with $x_2\gt x_1$.

Since $f'''(x) \gt 0$ for all $x \in \mathbb {R}$, $f''(x)$ is an increasing function. And in Taylor's expansion i will be ending at $f''(x)$ but not sure how to bring in $\frac{x_1+x_2}{2}$.

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2 Answers 2

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Using the Taylor expansion to third order, for all $y$ there exists $\zeta$ between $(x_1+x_2)/2$ and $y$ such that $$ f(y) = f \left( \frac{x_1+x_2}2 \right) + f'\left( \frac{x_1+x_2}2 \right)\left(y - \frac{x_1 + x_2}2 \right) \\ + \frac{f''(\frac{x_1+x_2}2)}2 \left( y - \frac{x_1 + x_2}2 \right)^2 + \frac{f'''(\zeta)}6 \left( y - \frac{x_1+x_2}2 \right)^3. \\ $$ It follows that by plugging in $x_2$, $$ f(x_2) - f \left( \frac{x_1+x_2}2 \right) > f' \left( \frac{x_1+x_2}2 \right) \frac {x_2-x_1}2 + \frac{f''(\frac{x_1+x_2}2)}2 \left(\frac{x_2-x_1}2 \right)^2 $$ since $f'''(\zeta) > 0$ and $x_2 > \frac{x_1+x_2}2$. Similarly, by plugging in $x_1$, $$ f(x_1) - f \left( \frac{x_1+x_2}2 \right) < f' \left( \frac{x_1+x_2}2 \right) \frac {x_1-x_2}2 + \frac{f''(\frac{x_1+x_2}2)}2 \left(\frac{x_1-x_2}2 \right)^2, $$ (note that the sign that appears in the cubic term is now negative, hence the reversed inequality) which we can re-arrange as $$ f \left( \frac{x_1+x_2}2 \right) - f(x_1) > f' \left( \frac{x_1+x_2}2 \right) \frac {x_2-x_1}2 - \frac{f''(\frac{x_1+x_2}2)}2 \left(\frac{x_1-x_2}2 \right)^2, $$ By adding up, the quadratic terms cancel out and you get your result.

Hope that helps,

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Set $a = \frac{x_1+x_2}{2}$ and $x = \frac{x_2-x_1}{2}$. Then the claim is $$ f(a+x) - f(a-x) > 2xf'(a) $$ for $x > 0$. In order to prove this, apply Taylor's theorem to $$ g(x) = f(a+x) - f(a-x) $$ and note that $g(0) = 0$, $g'(0) =2 f'(a)$, $g''(0) = 0$, and $g'''(x) > 0$.

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