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\[ f(x) = -2x^3 + 6x - 2 \]
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.
![curve with a secant and two tangent: mean value theorem.](http://www.analyzemath.com/calculus/MeanValueTheorem/mean_value_1.gif)
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 2Function \( \cos x \) 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
DifferentiationMean Value Theorem