# Proving equivalent form of the mean-value theorem for derivatives

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.

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Take $a=x$ and $b=x+h$. Note "$c$ in $(a,b)$" is then "$x+\theta h$ for some $0<\theta<1$". – David Mitra Mar 15 '12 at 23:58
Thank you! Seeing $c \in (a,b) \implies c = x + \theta h$ was eluding me for some reason. I feel dumb as it seems painfully obvious now... If you wanted to leave that as answer, I could accept it. – user23784 Mar 16 '12 at 0:04

It's just notation. Fix $x$ and $h \ne0$. Let $a=x$ and $b=x+h$. The Mean Value Theorem states $${f(b)-f(a)}=f'(c)(b-a),\quad \text{for some }c \text{ between } a\text{ and }b..$$ Rewrite this slightly to obtain $${f(b)}=f(a)+f'(c)(b-a),\quad \text{for some }c \text{ between } a\text{ and }b.$$
Now write in terms of $x$ and $h$. Note here that if $c$ is between $x$ and $x+h$, then $c=x+\theta h$ for some $0<\theta<1$. Also $b-a=(x+h)-x=h$. So you get $$f(x+h)=f(x)+hf'(x+\theta h)\quad \text{for some } 0<\theta<1.$$
(Of course, you need to assume that $f$ is continuous and differentiable on the appropriate intervals.)