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Lagrange mean value theorem

9 bytes added, 20:18, 20 October 2011
Proof
| 4 || <math>g</math> is differentiable on <math>(a,b)</math> || Fact (2) || <math>f</math> is differentiable on <math>(a,b)</math> || Steps (1), (2) || <toggledisplay>By Step (2), <math>g = f - h</math>. By Step (1), <math>h</math> is linear and hence differentiable on <math>(a,b)</math>. Thus, by Fact (1), <math>g = f - h</math>, being the difference of two differentiable functions, is differentiable on <math>(a,b)</math>.</toggledisplay>
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| 5 || <math>\! g(a) = g(b) = 0</math> || || || Steps (1), (2) || <toggledisplay><math>h(a) = f(a)</math> yields <math>g(a) = 0</math>. <math>h(b) = f(b)</math> yields <math>g(b) = 0</math>.</toggledisplay>
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| 6 || There exists <math>c \in (a,b)</math> such that <math>\! g'(c) = 0</math>. || Fact (3) || || Steps (3), (4), (5) || <toggledisplay>Steps (3), (4), (5) together show that <math>g</math> satisfies the conditions of Rolle's theorem on the interval <math>[a,b]</math>, hence we get the conclusion of the theorem.</toggledisplay>
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| 7 || For the <math>c</math> obtained in step (6), <math>\! f'(c) = h'(c)</math> || Fact (4) || || Steps (2), (6) || <toggledisplay>From Step (2), <math>g = f - h</math>. Differentiating both sides by Fact (4), we get <math>g' = f' - h'</math> on <math>(a,b)</math>. Since <math>g'(c) = 0</math>, we obtain that <math>f'(c) = h'(c)</math>.</toggledisplay>
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| 8 || <math>h'(x) = \frac{f(b) - f(a)}{b - a}</math> for all <math>x</math>. In particular, <math>h'(c) = \frac{f(b) - f(a)}{b - a}</math>. || || || Step (1) || Differentiate the expression for <math>h(x)</math> from Step (1).
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