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Lagrange mean value theorem

2,997 bytes added, 20:12, 20 October 2011
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* [[Increasing and differentiable implies nonnegative derivative]]
* [[Derivative of differentiable function on interval satisfies intermediate value property]]
 
==Facts used==
 
# [[uses::Continuous functions form a vector space]]
# [[uses::Differentiable functions form a vector space]]
# [[uses::Rolle's theorem]]
# [[uses::Differentiation is linear]]
==Proof==
 
{| class="sortable" border="1"
! Step no. !! Assertion/construction !! Facts used !! Given data used !! Previous steps used !! Explanation
|-
| 1 || Consider the function <math>h(x) := \frac{f(b) - f(a)}{b - a}\cdot x + \frac{bf(a) - af(b)}{b - a}</math>. Then, <math>h</math> is a linear (and hence a continuous and differentiable) function with <math>h(a) = f(a)</math> and <math>h(b) = f(b)</math>|| || || Just plug in and check. Secretly, we obtained <math>h</math> by trying to write the equation of the line joining the points <math>(a,f(a))</math> and <math>(b,f(b))</math>.
|-
| 2 || Define <math>g = f - h</math> on <math>[a,b]</math>, i.e., <math>g(x) := f(x) - h(x)</math>. || || ||
|-
| 3 || <math>g</math> is continuous on <math>[a,b]</math> || Fact (1) || <math>f</math> is continuous 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 continuous on <math>[a,b]</math>. Thus, by Fact (1), <math>g = f - h</math>, being the difference of two continuous functions, is continuous on <math>[a,b]</math>.</toggledisplay>
|-
| 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>
|-
| 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>
|-
| 7 || For the <math>c</math> obtained in step (6), <math>f'(c) = h'(c)</math> || Fact (4) || || Steps (2), (6) || 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>
|-
| 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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| 9 || <math>f'(c) = \frac{f(b) - f(a)}{b - a}</math> || || || Steps (7), (8) || Step-combination direct
|}
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