I was given the following question as part of my real analysis studies, and I'm fairly certain the question is impossible. Prove the following: Let f be a differentiable function defined on [a,b]. Then for all t in the reals, such that f'(a)<t<f'(b) there exists x∈(a,b) such that f'(x)=t.
The reason I'm struggling is this seems to be a mean value theorem related problem, however without the condition that f is twice differentiable or f' is continuous or some other similar condition, I believe that t does not have to exist by virtue of derivatives of functions not having to be continuous.
Help is appreciated or confirmation that the wording is wrong somehow.
This is not a "mean value theorem" but an "intermediate value theorem", which is a well-known fact in basic analysis.
If necessary, replace $f(x)$ by $f(x)-tx$, and thus we may assume $t=0$. Suppose $f(x)$ attains a minimum on the interval $[a, b]$ at $x=x_0$ (which exists as $f(x)$ is continuous), and we have $f'(x_0)=0$.