# Difference between revisions of "Lagrange mean value theorem"

From Calculus

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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>. | + | | 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

## Contents

## Statement

Suppose is a function defined on a closed interval (with ) such that the following two conditions hold:

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

Then, there exists in the open interval such that the derivative of at equals the difference quotient . More explicitly:

Geometrically, this is equivalent to stating that the tangent line to the graph of at is parallel to the chord joining the points and .

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

## Related facts

- Rolle's theorem
- Zero derivative implies locally constant
- Fundamental theorem of calculus
- Positive derivative implies increasing
- Increasing and differentiable implies nonnegative derivative
- Derivative of differentiable function on interval satisfies intermediate value property

## Facts used

- Continuous functions form a vector space
- Differentiable functions form a vector space
- Rolle's theorem
- Differentiation is linear

## Proof

Step no. | Assertion/construction | Facts used | Given data used | Previous steps used | Explanation |
---|---|---|---|---|---|

1 | Consider the function . Then, is a linear (and hence a continuous and differentiable) function with and | Just plug in and check. Secretly, we obtained by trying to write the equation of the line joining the points and . | |||

2 | Define on , i.e., . | ||||

3 | is continuous on | Fact (1) | is continuous on | Steps (1), (2) | [SHOW MORE] |

4 | is differentiable on | Fact (2) | is differentiable on | Steps (1), (2) | [SHOW MORE] |

5 | Steps (1), (2) | [SHOW MORE] | |||

6 | There exists such that . | Fact (3) | Steps (3), (4), (5) | [SHOW MORE] | |

7 | For the obtained in step (6), | Fact (4) | Steps (2), (6) | From Step (2), . Differentiating both sides by Fact (4), we get on . Since , we obtain that .</toggledisplay> | |

8 | for all . In particular, . | Step (1) | Differentiate the expression for from Step (1). | ||

9 | Steps (7), (8) | Step-combination direct |