Page:Elements of the Differential and Integral Calculus - Granville - Revised.djvu/189

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106. The Theorem of Mean Value.^[1] Consider the quantity Q defined by the equation

(A)	${\begin{smallmatrix}{\frac {f(b)-f(a)}{b-a}}=Q,\end{smallmatrix}}$ or
(B)	$f(b)-f(a)-(b-a)Q=0.$
Let $F(x)$ be a function formed by replacing b by x in the left-hand member of (B); that is,
(C)	$F(x)=f(x)-f(a)-(x-a)Q$ .
From (B),	$F(b)=0$ , and from (C), $F(a)=0$ ;
therefore, by Rolle's Theorem (§105) $F'(x)$ must be zero for at least one value of x between a and b, say for $x_{1}$ . But by differentiating (C) we get
	$F'(x)=f'(x)-Q.$
Therefore, since	$F'(x_{1})=0,$ then also $f'(x_{1})-Q=0$ ,
and	$Q=f'(x_{1}).$
Substituting this value of Q in (A), we get the Theorem of Mean Value,
(44)	${\begin{smallmatrix}{\frac {f(b)-f(a)}{b-a}}=f'(x_{1}),\ a<x_{1}<b\end{smallmatrix}}$

where in general all we know about $x_{1}$ is that it lies between a and b.

The Theorem of Mean Value interpreted Geometrically. Let the curve in the figure be the locus of

y=f(x).

Take OC = a and OD = b; then $f(a)=CA$ and $f(b)=DB$ , giving $AE=b-a$ and $EB=f(b)-f(a)$ .

Therefore the slope of the chord AB is

(D)	${\begin{smallmatrix}\tan EAB={\frac {EB}{AE}}={\frac {f(b)-f(a)}{b-a}}.\end{smallmatrix}}$
There is at least one point on the curve between A and B (as P) where the tangent (or curve) is parallel to the chord AB. If the abscissa of P is Xl' the slope at P is
(E)	$\tan t=f'(x_{1})=\tan EAB.$