# Mean Value Theorem Problems

Problems related to the mean value theorem, with detailed solutions, are presented.

## Mean Value Theorem: Review

If f is a function continuous on the interval [ a , b ] and differentiable on (a , b ), then at least one real number c exists in the interval (a , b) such that
f '(c) = [f(b) - f(a)] / (b - a).

The mean value theorem expresses the relatonship between the slope of the tangent to the curve at x = c and the slope of the secant to the curve through the points (a , f(a)) and (b , f(b)).

### Problem 1

Find a value of c such that the conclusion of the mean value theorem is satisfied for
f(x) = -2x 3 + 6x - 2

on the interval [-2 , 2]

Solution to Problem 1
f(x) is a polynomial function and is continuous and differentiable for all real numbers. Let us evalute f(x) at x = -2 and x = 2
f(-2) = -2(-2) 3 + 6(-2) - 2 = 2
f(2) = -2(2) 3 + 6(2) - 2 = - 6
Evaluate [f(b) - f(a)] / (b - a)
[f(b) - f(a)] / (b - a) = [ -6 - 2 ] / (2 - -2) = -2
Let us now find f '(x).
f '(x) = - 6x 2 + 6
We now construct an equation based on f '(c) = [f(b) - f(a)] / (b - a)
-6c 2 + 6 = -2
Solve for c to obtain 2 solutions
c = 2 √(1/3) and c = - 2 √(1/3)
Below is shown the graph of f, a secant and the two tangents corresponding to the two solutions found. The secant and the two tangents are parallel since their slopes are equal according to the mean value theorem. ### Problem 2

Use the mean value theorem to prove that for any two real numbers a and b,
| cos a - cos b| ≤ | a - b|

Solution to Problem 2
Function cos x is continuous and differentiable for all real numbers. Use the mean value theorem, using 2 real numbers a and b to write
(cos x) ' = (cos a - cos b) / (a - b)
Take the absolute value of both sides
| (cos x) ' | = | (cos a - cos b) / (a - b) |
(cos x)' = - sin x, hence.
| (cos x) ' | ≤ 1
Which gives
| (cos a - cos b) / (a - b) | ≤ 1
But
| (cos a - cos b) / (a - b) | = |cos a - cos b| / |a - b|
When combined with the above gives
|cos a - cos b| / |a - b| ≤ 1
Multiply both sides by |a - b| to obtain
|cos a - cos b| ≤ |a - b|