I'm trying to prove the following statement (from Apostol's Calculus I, p. 186, Exercise 6)
Show that the mean-value formula can be expressed in the form f(x+h) = f(x) + hf'(x + \theta h) \qquad \text{where} \, 0 < \theta < 1.
As I'm not certain how standard the presentation of the mean-value formula is, Apostol gives it as f(b) - f(a) = f'(c)(b-a) For $f$ continuous on $[a,b]$ and having a derivative at each point of the open interval $(a,b)$ where $c \in (a,b)$.
This really does not strike me as a problem that should be too difficult, but for some reason I can't seem to make the right connections.
First, I notice that f(x+h) = f(x) + hf'(x + \theta h) \implies \frac{f(x+h) - f(x)}{h} = f'(x + \theta h) looks like we are getting something reminiscent of the derivative of $f$ at $x$ on the left. It's not clear to me what I can or should do with that.
The other thing that it occurs to me to do, is to write $x = \theta x + (1-\theta)x$ for $0 < \theta < 1$. I'm not sure what to do with that precisely, but it seems like the sort of thing that one should do in this situation.
I wish I had more of my own work to show, but I can't seem to make any legitimate progress.