# 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 thatThe 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 - 2on 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|