Show that if $f'(a)
Let $f:I\rightarrow \mathbb R$ be a differentiable function on the
  open interval $I$. Show that if $f'(a)<y<f'(b)$ for some $a<b$ in $I$
  and $y\in \mathbb R$, then there exists $x\in I$ with $a<x<b$ and
  $f'(x)=y.$ [Note that $f'$ is not assumed to be continuous.]
I can't see why the question has specified $f'$. What I mean is, why can't we set $g(x)=f'(x)$? We then still have that $g:I\rightarrow \mathbb R$ and everything that applies to $f$ applies to $g$ (except that $g$ is not necessarily continuous). Are we not then trying to find $g(x)=y$ where $g(a)<y<g(b)$? Why did the question not simple ask this of $f$ and simply omit the detail that $f$ is differentiable? Hopefully this query makes some sense.
Regardless I cannot do the question. We could simply apply the Intermediate Value Theorem if it were given that $f'$ is continuous, so I have thought about somehow showing that $f'$ is continuous but I don't think that's right. I have also tried splitting $f'$ up into continuous blocks and applied the Intermediate Value Theorem but that doesn't seem to be getting me anywhere.
I am asking for help in understanding why it has to be a derivative but also would like some assistance on the question.
Any help is appreciated, 
Thank you.
 A: Hint: replacing $f$ by $f_1 : x\mapsto f(x)-yx$, we can assume $y=0$. Now we're searching for a solution of the equation $f'(x)=0$ and you may try to search for an extremum of $f$ on the closed bounded interval $[a,b]$. 
Note that I don't use a continuity hypothesis on $f'$ because $f'$ need not be continuous but there are discontinuous functions which can't be the derivative of any function (precisely because the statement of your exercise is false for them...). 
A: Because not all functions that you could put in the role of $g$ are derivatives.  For example, suppose $$g(x) = \begin{cases} 0 & \text{for }x< 0, \\ 1 & \text{for } x\ge0, \end{cases}$$ and the open interval $I$ contains $0$.  Then there is no function $f$ such that $f'=g$ everywhere in $I$.  And $g$ does not assume any values between $0$ and $1$, so the conclusion you're trying to prove does not hold.
The result is called Darboux's theorem: https://en.wikipedia.org/wiki/Darboux%27s_theorem_(analysis)
Here's the idea of what I suspect is the most usual proof: First reduce the problem to that of proving the result in the case where $y=0$. The way to do that is similar to the way you reduce the problem of proving the mean value theorem to the problem of proving Rolle's theorem.  Then show that since $f$ is continuous, it has to assume a maximum and a minimum in the closed interval from $a$ to $b$.  Then think about why at least one of the extreme values cannot be at an endpoint of that interval.  And recall that where $f$ has an extreme value, $g$ must be $0$.
That a function can be a derivative at every point without being continuous is perhaps counterintuitive.  An example is the derivative of a function satisfying $-x^2 \le f(x) \le x^2$, where $f'(x)=1$ for $x = 1/(2n)$ and $f'(x)=-1$ for $x$ somewhere between two of those points.  Since $f'$ oscillates between $1$ and $-1$ as $x\to 0$, $f'$ cannot be continuous at $0$.  But because $-x^2\le f(x)\le x^2$, one can show that $f'(0)=0$, so $f'$ is everywhere defined even though it has a discontinuity.
