Problems related to the mean value theorem, with detailed solutions, are presented.
In what follows, \( C \) represents the constant of integration where applicable.
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 relationship 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 \( (b, f(b)) \).
Problems with Detailed Solutions
Click on each problem to view its detailed step-by-step solution.
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 \( [-2, 2] \).
Show Solution to Problem 1
\( f(x) \) is a polynomial function and is continuous and differentiable for all real numbers. Let us evaluate \( f(x) \) 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 \( \dfrac{f(b) - f(a)}{b - a} \):
\[ \dfrac{f(b) - f(a)}{b - a} = \dfrac{-6 - 2}{2 - (-2)} = \dfrac{-8}{4} = -2 \]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:
\[ -6c^2 = -8 \implies c^2 = \dfrac{8}{6} = \dfrac{4}{3} \] \[ c = \pm \sqrt{\dfrac{4}{3}} = \pm \dfrac{2}{\sqrt{3}} = \pm \dfrac{2\sqrt{3}}{3} \]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 \( b \):
\[ |\cos a - \cos b| \le |a - b| \]Show Solution to Problem 2
The function \( f(x) = \cos x \) is continuous and differentiable for all real numbers. Applying the Mean Value Theorem on the interval between \( a \) and \( b \) (assuming \( a \neq b \)):
\[ \dfrac{\cos a - \cos b}{a - b} = f'(c) \]where \( c \) is between \( a \) and \( b \). Since \( f'(x) = -\sin x \), we have:
\[ \dfrac{\cos a - \cos b}{a - b} = -\sin c \]Take the absolute value of both sides:
\[ \left| \dfrac{\cos a - \cos b}{a - b} \right| = |-\sin c| = |\sin c| \]Since \( |\sin c| \le 1 \) for any real number \( c \), we have:
\[ \left| \dfrac{\cos a - \cos b}{a - b} \right| \le 1 \]From the properties of absolute value, \( \left| \dfrac{x}{y} \right| = \dfrac{|x|}{|y|} \):
\[ \dfrac{|\cos a - \cos b|}{|a - b|} \le 1 \]Multiply both sides by \( |a - b| \) to obtain:
\[ |\cos a - \cos b| \le |a - b| \]Note: If \( a = b \), the inequality trivially holds as \( 0 \le 0 \).