Editing Mean value theorem (section)

== Statement ==
[[File:Mittelwertsatz3.svg|thumb|The function <math>f</math> attains the slope of the secant between <math>a</math> and <math>b</math> as the derivative at the point <math>\xi\in(a,b)</math>.]]
[[File:Mittelwertsatz6.svg|thumb|It is also possible that there are multiple tangents parallel to the secant.]]
Let <math>f:[a,b]\to\R</math> be a [[continuous function]] on the [[closed interval]] {{nowrap|<math>[a,b]</math>,}} and [[differentiable function|differentiable]] on the [[open interval]] {{nowrap|<math>(a,b)</math>,}} where {{nowrap|<math>a < b</math>.}} Then there exists some <math>c</math> in <math>(a,b)</math> such that:{{sfn|Rudin|1976|p=108}}

:<math>f'(c)=\frac{f(b)-f(a)}{b-a}.</math>

The mean value theorem is a generalization of [[Rolle's theorem]], which assumes <math>f(a)=f(b)</math>, so that the right-hand side above is zero.

The mean value theorem is still valid in a slightly more general setting. One only needs to assume that <math>f:[a,b]\to\R</math> is [[continuous function|continuous]] on <math>[a,b]</math>, and that for every <math>x</math> in <math>(a,b)</math> the [[limit of a function|limit]]

:<math>\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}</math>

exists as a finite number or equals <math>\infty</math> or <math>-\infty</math>. If finite, that limit equals <math>f'(x)</math>. An example where this version of the theorem applies is given by the real-valued [[cube root]] function mapping <math>x \mapsto x^{1/3}</math>, whose [[derivative]] tends to infinity at the origin.

===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>