understanding IVT vs MVT I know that if the MVT was applied to physics then it would roughly translate to saying that the average velocity = instantaneous velocity.
But suppose that my average velocity on $[0,T]$ was $10$. Then  $\frac{f(T)-f(0)}{T-0}=10.$ Assuming that $\int f' \,dt = f$, then $\frac{f(T)-f(0)}{T-0} = \frac{\int^{T}_{0} f' \,dt}{T}=10.$
In other words, the average value of $f'(t)$ is $10$ on $[0,T]$. Intuitively, since $f'(t)$ is continuous, if we interpret it as a continuous velocity function, then $f'(t)$ has to take values greater than 10 and less than 10. But then I can apply the IVT to say that $\exists \, c$ s.t. $f'(c)=10$, i.e. instantaneous velocity at $c$ is 10 or the average velocity...
This confuses me because it appears that the MVT and IVT can be both applied to solve this. any thoughts?

Hopefully this isn't awkward, but following anorton's suggestion I've copied my comment to an answer:
That is correct. If $f′(t)$ is continuous, then you can apply the IVT as you do. Lots of books don't require this hypothesis for the MVT, though. All you need for the MVT is $f$ continuous on $[0,T]$ and differentiable on $(0,T)$. It is possible to apply the MVT even if $f′(t)$ is not continuous on $(0,T)$.
