Editing Mean value theorem (section)

===Proof===
The expression <math display="inline">\frac{f(b)-f(a)}{b-a}</math> gives the [[slope]] of the line joining the points <math>(a,f(a))</math> and <math>(b,f(b))</math>, which is a [[Chord (geometry)|chord]] of the graph of <math>f</math>, while <math>f'(x)</math> gives the slope of the tangent to the curve at the point <math>(x,f(x))</math>. Thus the mean value theorem says that given any chord of a smooth curve, we can find a point on the curve lying between the end-points of the chord such that the tangent of the curve at that point is parallel to the chord. The following proof illustrates this idea.

Define <math>g(x)=f(x)-rx</math>, where <math>r</math> is a constant. Since <math>f</math> is continuous on <math>[a,b]</math> and differentiable on <math>(a,b)</math>, the same is true for <math>g</math>. We now want to choose <math>r</math> so that <math>g</math> satisfies the conditions of [[Rolle's theorem]]. Namely

:<math>\begin{align}
g(a)=g(b)&\iff f(a)-ra=f(b)-rb\\
&\iff r(b-a)=f(b)-f(a) \\
&\iff r=\frac{f(b)-f(a)}{b-a} .
\end{align}</math>

By [[Rolle's theorem]], since <math>g</math> is differentiable and <math>g(a)=g(b)</math>, there is some <math>c</math> in <math>(a,b)</math> for which <math>g'(c)=0</math> , and it follows from the equality <math>g(x)=f(x)-rx</math> that,

:<math>\begin{align}
&g'(x) = f'(x) -r \\
& g'(c) = 0\\
&g'(c) = f'(c) - r = 0 \\
&\Rightarrow f'(c) = r = \frac{f(b)-f(a)}{b-a}
\end{align}</math>