# Mean Value Theorem Problems

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

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

\( f'(c) = \dfrac{{f(b) - f(a)}}{{b - a}} \).

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 \( \dfrac{{f(b) - f(a)}}{{b - a}} \)

\( \dfrac{{f(b) - f(a)}}{{b - a}} = \dfrac{{-6 - 2}}{{2 - (-2)}} = -2 \)

Let us now find \( f'(x) \).

\( f'(x) = - 6x^2 + 6 \)

We now construct an equation based on \( f'(c) = \dfrac{{f(b) - f(a)}}{{b - a}} \)

\( -6c^2 + 6 = -2 \)

Solve for \( c \) to obtain 2 solutions

\( c = 2 \sqrt{\dfrac{1}{3}} \) and \( c = - 2 \sqrt{\dfrac{1}{3}} \)

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
\[ (\cos x)' = \dfrac{{\cos a - \cos b}}{{a - b}} \]
Take the absolute value of both sides
\[ | (\cos x)' | = \left| \dfrac{{\cos a - \cos b}}{{a - b}} \right| \]
\( (\cos x)' = - \sin x \), and therefore \( | (\cos x)' | ≤ 1 \)

Which gives
\[ \left| \dfrac{{\cos a - \cos b}}{{a - b}} \right| ≤ 1 \]
From the properties of absolute value, we have
\[ \left| \dfrac{\cos a - \cos b}{a-b} \right| = \dfrac{|\cos a - \cos b|} { |a - b|} \]
When combined with the above gives
\[ \dfrac {| \cos a - \cos b |}{| a - b |} ≤ 1 \]
Multiply both sides by | a - b | to obtain
\[ |\cos a - \cos b | ≤ | a - b | \]

## More references

Differentiation

Mean Value Theorem