Lagrange mean value theorem

From Calculus
Jump to: navigation, search


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


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