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

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