The mean value theorem is one of the most important theorems in calculus. It is here discussed through examples and graphs. Once you finish with this tutorial you might want to solve problems related to the mean value theorem.
In what follows, \( C \) represents the constant of integration where applicable.
Mean Value Theorem
Let \( f(x) \) be a continuous function on the closed interval \( [a, b] \) and differentiable on the open interval \( (a, b) \). Then there is at least one value \( c \) of \( x \) in the interval \( (a, b) \) such that:
\[ f'(c) = \dfrac{f(b) - f(a)}{b - a} \]or
\[ f(b) - f(a) = f'(c)(b - a) \]In other words, the tangent line to the graph of \( f \) at \( c \) and the secant through points \( (a, f(a)) \) and \( (b, f(b)) \) have equal slopes and are therefore parallel.
Examples on the Applications of the Mean Value Theorem
Click on each example to view its detailed step-by-step solution.
Example 1
Use the mean value theorem to find the value \( c \) of \( x \) in the interval \( [1, 5] \) such that the tangent at the point \( (c, f(c)) \) to the curve \( f(x) = -x^2 + 7x - 6 \) is parallel to the secant through the points \( (1, f(1)) \) and \( (5, f(5)) \).
Show Solution to Example 1
The slope of the tangent at point \( (c, f(c)) \) is given by:
\[ f'(c) \]where \( f' \) is the first derivative. The slope of the secant through \( (1, f(1)) \) and \( (5, f(5)) \) is given by:
\[ \dfrac{f(5) - f(1)}{5 - 1} \]For the tangent to be parallel to the secant, their slopes have to be equal, hence:
\[ f'(c) = \dfrac{f(5) - f(1)}{5 - 1} \]Function \( f \) is a polynomial (quadratic) function and is therefore continuous and differentiable on the interval \( [1, 5] \), so the Mean Value Theorem predicts that there is at least one value of \( x (= c) \) such that the above equality is true.
The first derivative is:
\[ f'(x) = -2x + 7 \]The slope \( m_1 \) of the tangent to the curve at \( x = c \) is:
\[ m_1 = f'(c) = -2c + 7 \]The slope \( m_2 \) of the secant through the points \( (1, f(1)) \) and \( (5, f(5)) \) is:
\[ m_2 = \dfrac{f(5) - f(1)}{5 - 1} = \dfrac{4 - 0}{4} = 1 \]Setting \( m_1 = m_2 \) gives the equation:
\[ -2c + 7 = 1 \implies -2c = -6 \implies c = 3 \]Check answer graphically:
Point of tangency at \( x = c \) is given by \( (3, f(3)) = (3, 6) \).
Equation of tangent:
\[ y - 6 = 1(x - 3) \implies y = x + 3 \]In Figure 2 below are shown the graphs of the given function and the graph of the tangent to the curve of \( f \). The tangent and secant have equal slopes and are therefore parallel.
Note: There may be more than one value of \( x (= c) \) that satisfies the mean value theorem, as seen in Example 2 below.
Example 2
Use the mean value theorem to find all values of \( x \) in the interval \( [0, 3] \) such that the tangent at the points \( (c, f(c)) \) to the curve \( f(x) = x^3 - 5x^2 + 7x + 1 \) is parallel to the secant through the points \( (0, f(0)) \) and \( (3, f(3)) \).
Show Solution to Example 2
Function \( f \) is a polynomial function and is therefore continuous and differentiable on the interval \( [0, 3] \), so the Mean Value Theorem predicts that there is at least one value of \( x (= c) \) such that the tangent to the curve of \( f \) at \( x = c \) and the secant are parallel and their slopes are equal.
The first derivative is:
\[ f'(x) = 3x^2 - 10x + 7 \]The slope \( m_1 \) of the tangent at \( x = c \) is:
\[ m_1 = f'(c) = 3c^2 - 10c + 7 \]The slope \( m_2 \) of the secant through the points \( (0, f(0)) \) and \( (3, f(3)) \) is:
\[ m_2 = \dfrac{f(3) - f(0)}{3 - 0} = \dfrac{4 - 1}{3} = \dfrac{3}{3} = 1 \]For the tangent to the curve at \( x = c \) and the secant through \( (0, f(0)) \) and \( (3, f(3)) \) to be parallel, their slopes must be equal:
\[ 3c^2 - 10c + 7 = 1 \]Which may be written as:
\[ 3c^2 - 10c + 6 = 0 \]Solve using the quadratic formula to obtain two solutions:
\[ c = \dfrac{-(-10) \pm \sqrt{(-10)^2 - 4(3)(6)}}{2(3)} = \dfrac{10 \pm \sqrt{100 - 72}}{6} = \dfrac{10 \pm \sqrt{28}}{6} = \dfrac{10 \pm 2\sqrt{7}}{6} = \dfrac{5 \pm \sqrt{7}}{3} \]Calculating the decimal approximations:
\[ c_1 = \dfrac{5 - \sqrt{7}}{3} \approx 0.78 \quad \text{and} \quad c_2 = \dfrac{5 + \sqrt{7}}{3} \approx 2.55 \]Check answer graphically:
In Figure 3 below are shown the graphs of the given function and the graph of the two tangents to the curve of \( f \) parallel to the secant through the points \( A(0, f(0)) \) and \( B(3, f(3)) \).
