Suppose $f$ is a thrice differentiable function on $\mathbb {R}$ . Showing an identity using taylor's theorem 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 $\frac{x_1+x_2}{2}$. 

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 $g'''(x)  > 0$.
