Difference between revisions of "Lagrange mean value theorem"

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(Proof)
(Proof)
 
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| 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>
 
| 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>
+
| 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>
 
| 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>

Latest revision as of 20:19, 20 October 2011

Statement

Suppose f is a function defined on a closed interval [a,b] (with a < b) such that the following two conditions hold:

  1. f is a continuous function on the closed interval [a,b] (i.e., it is right continuous at a, left continuous at b, and two-sided continuous at all points in the open interval (a,b)).
  2. f is a differentiable function on the open interval (a,b), i.e., the derivative exists at all points in (a,b). Note that we do not require the derivative of f to be a continuous function.

Then, there exists c in the open interval (a,b) such that the derivative of f at c equals the difference quotient \Delta f(a,b). More explicitly:

f'(c) = \frac{f(b) - f(a)}{b - a}

Geometrically, this is equivalent to stating that the tangent line to the graph of f at c is parallel to the chord joining the points (a,f(a)) and (b,f(b)).

Note that the theorem simply guarantees the existence of c, and does not give a formula for finding such a c (which may or may not be unique).

Related facts

Facts used

  1. Continuous functions form a vector space
  2. Differentiable functions form a vector space
  3. Rolle's theorem
  4. Differentiation is linear

Proof

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 Consider the function
h(x) := \frac{f(b) - f(a)}{b - a} x +\frac{bf(a) - af(b)}{b - a}.
Then, h is a linear (and hence a continuous and differentiable) function with h(a) = f(a) and h(b) = f(b)
Just plug in and check. Secretly, we obtained h by trying to write the equation of the line joining the points (a,f(a)) and (b,f(b)).
2 Define g = f - h on [a,b], i.e., g(x) := f(x) - h(x).
3 g is continuous on [a,b] Fact (1) f is continuous on [a,b] Steps (1), (2) [SHOW MORE]
4 g is differentiable on (a,b) Fact (2) f is differentiable on (a,b) Steps (1), (2) [SHOW MORE]
5 g(a) = g(b) = 0 Steps (1), (2) [SHOW MORE]
6 There exists c \in (a,b) such that \!g'(c) = 0. Fact (3) Steps (3), (4), (5) [SHOW MORE]
7 For the c obtained in step (6), \! f'(c) = h'(c) Fact (4) Steps (2), (6) [SHOW MORE]
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).
9 f'(c) = \frac{f(b) - f(a)}{b - a} Steps (7), (8) Step-combination direct