Difference between revisions of "Lagrange mean value theorem"

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(Proof)
(Proof)
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! Step no. !! Assertion/construction !! Facts used !! Given data used !! Previous steps used !! Explanation
 
! Step no. !! Assertion/construction !! Facts used !! Given data used !! Previous steps used !! Explanation
 
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| 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>.
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| 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>.
 
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| 2 || Define <math>g = f - h</math> on <math>[a,b]</math>, i.e., <math>g(x) := f(x) - h(x)</math>. || || || ||
 
| 2 || Define <math>g = f - h</math> on <math>[a,b]</math>, i.e., <math>g(x) := f(x) - h(x)</math>. || || || ||

Revision as of 20:13, 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}\cdot 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) 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>
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