# Changes

## Lagrange mean value theorem

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