Using Mean Value Theorem to Prove Derivative Greater than Zero I'm working on a problem where at one point I have to show that for $x\ge a$,  
$$g (x) = \int_a^x f - (x-a) f \left({a+x \over 2} \right)$$, $g'(x) \ge 0$. 
Additional information: I know that $f''(x)\gt0$, $f'(x)<0$, and $f(x)\gt0$. 
Here is what I have so far:
$g'(x) = f(x) - f(a) - f({a+x \over 2}) - (x-a) f'({a+x \over 2})({1 \over 2})$
Using the Mean Value Theorem, I replaced $f(x) - f(a)$ with $(x-a) f'(x_0)$ where $x_0 \in [a,x]$, resulting in: $\displaystyle g'(x) = (x-a)\left[f'(x_0) - {1 \over 2} f'\left({a+x \over 2}\right)\right] - f\left({a+x \over 2}\right)$.
I am having trouble, however, going from this step (assuming I've taken the right steps thus far, and getting the result that $g'(x) \ge 0$.

Some hints/corrections:


*

*The derivative of $\int_a^x f$ is just $f(x)$.

*The way you used the mean value theorem in your above attempt is absolutely correct. You'll need to apply it twice, which you might expect since you were given information about $f''$.

