Calculus: Prove function is increasing This question is from Stewart's Essential Calculus:

Suppose $f$ is differentiable on an interval $I$ and $f'(x) > 0$ for all 
  numbers $x$ in $I$ except for a single number $c$. Prove that $f$ is 
  increasing on the entire interval $I$.

The main thrust of Stewart's proof (in his student solutions manual) is considering $x_1 < c < x_2$ and showing that $f(x_1) < f(c) < f(x_2)$, which he handles piecemeal with the Mean Value Theorem. So for the left-hand inequality, he shows that $f(c) - f(x_1) > 0$ because $c - x_1 > 0$ and $f'(k) > 0$ for some $k$ in $(x_1, c)$. I understand that.
I just want to know whether my clumsier answer using the Intermediate Value Theorem is at least correct (vs. some gross error for which I'll slap myself or some analytic subtlety that's beyond me at this point). So for the left-hand inequality, wanting to show that $f(x_1) < f(c)$, I assume $f(x_1) >= f(c)$ and find a contradiction. 
CASE A: $f(x_1) > f(c)$. By the Intermediate Value Theorem, there's a $k$ in $(x_1, c)$ such that $f(x_1)>f(k)$, contradicting $f'(x)>0$ over $[x_1, k]$. 
CASE B: $f(x_1)=f(c)$. For all $m$ in $(x_1, c)$, $f(x_1) >= f(m)$ contradicts $f'(x) > 0$ over $[x_1, m]$, whereas $f(x_1) < f(m)$ implies $f(m) > f(c)$, falling to similar reasoning as in CASE A.
Thanks!
Edit: Apologies for the lack of MathJaX -- I haven't used it before. I did look up how to create subscripts, but it seems my underscores aren't doing anything useful. :) As for homework/revision, neither really -- this is just some self-study outside my main discipline. I got the student solutions manual so I could check my answers for all the odd-numbered exercises, but the solutions manual doesn't help with cases of "OK, but does this alternative answer work?"
 A: Your reasoning is correct. In both cases, you manage to find an interval away from $c$ on which  to apply the mean value theorem (MVT). 
That said, there isn't a reason to avoid $c$ being an endpoint of the interval on which MVT is applied. The statement of MVT involves derivative at some interior point of the interval. As long as you have control of $f'$ at the interior points, you can go ahead with the MVT.
A: We are given that $f$ is differentiable on $I$ and $f'(x) > 0$ for all $ x \in I - \{c\}$ where $c$ is some point in $I$. If $f'(c) < 0$ then by intermediate value property of derivatives there will be many points at which $f'(x) < 0$ and this is not allowed hence we must have $f'(c) = 0$.
Next let $a, b \in I$ with $a < b$. If $c \notin (a, b)$ then it is clear that $f(b) - f(a) = (b - a)f'(\xi) > 0$ where $\xi$ is some point in $(a, b)$. If $c \in (a, b)$ then we can see that $$f(b) - f(a) = f(b) - f(c) + f(c) - f(a) = (b - c)f'(\xi_{1}) + (c - a)f'(\xi_{2}) > 0$$ as each term is positive. Thus in all cases we have $f(a) < f(b)$ and therefore $f$ is strictly increasing in $I$.
I am not sure why you want to use IVT instead of MVT. Proving monotone nature of $f$ based on sign of derivative can be done by using IVT instead of MVT but not in the way you are doing. Rather in that case we need to augment IVT with Rolle's Theorem as shown in proof below.
Theorem: If $f$ is continuous on $[a, b]$ and $f'$ is positive on $(a, b)$ then $f$ is strictly increasing on $[a, b]$.
Proof without MVT: Let $a < c < d < b$ and we will show that $f(c) < f(d)$. Clearly $f(c) \neq f(d)$ otherwise $f'$ will vanish by virtue of Rolle's Theorem. If $f(c) > f(d)$ then $f'(c) > 0$ implies that there is a point $c' \in (c, d)$ such that $f(c') > f(c)$. Thus $f(d) < f(c) < f(c')$ and hence by IVT there is a point $d' \in (c', d)$ such that $f(c) = f(d')$ and this is again forbidden by Rolle's Theorem. Thus we must have $f(c) < f(d)$.
Next we show that $f(a) < f(d)$. By what we have proved above if $c', c, d$ are points with $a < c' < c < d < b$ then $f(c') < f(c) < f(d)$ and letting $c' \to a^{+}$ we can see that $f(a) \leq f(c) < f(d)$. Similarly we can show that $f(c) < f(b)$. Thus it follows that $f$ is strictly increasing on $[a, b]$.
