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.
Mean Value Theorem
Let \( f(x) \) be a continuous function on the 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.
Figure 1. Graphical Meaning of Mean Value Theorem
Examples on the Applications of the Mean Value Theorem
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 of curve \( f(x) = -x^2 + 7x - 6 \) is parallel to the secant through the points \( (1 , f(1)) \) and \( (5 , f(5)) \).
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 slope 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 of the interval \( [1 , 5] \) hence the mean value theorem predicts that there is a least one value of \( x (= c) \) such that the above equality is true.
The slope of the tangent is given by the value of the first derivative at \( x = c \).
The first derivative : \( f ' (x) = - 2x + 7 \)
slope \( m_1 \) of the tangent to the curve at x = c is equal to \[ m_1 = f'(c) = - 2c + 7 \]
The slope \( m_2 \) of the secant through the points \( (1 , f(1)) \) and \( (5 , f(5)) \) is given by
\[ m_2 = \dfrac{{f(5) - f(1)}}{{5 - 1}} = \dfrac{{4 - 0}}{4} = 1 \]
\( m_1 = m_2 \) gives the equation
\[ - 2c + 7 = 1 \]
\[ c = 3 \]
Check answer graphically
Point of tangency at \( x = c \) is given by \( (3 , f(3)) = (3 , 6) \)
Equation of tangent:
\[ y - 6 = (x - 3) \]
\[ y = x + 3 \]
In figure 1 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.
Figure 2. Mean Value Theorem used in example 1
There may be more than one value of \( x ( = c) \) that satisfies the mean value theorem, see 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 of curve \( f(x) = x^3 - 5x^2 + 7x + 1 \) is parallel to the secant through the points \( (0 , f(0)) \) and \( (3 , f(3)) \).
Solution to Example 2
Function f is a polynomial function and is therefore continuous and differentiable of the interval [1 , 3] and therefore 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 therefore their slopes are equal.
slope of tangent
The first derivative : \( f ' (x) = 3x^2 - 10x + 7 \)
The slope \( m_1 \) of the tangent at x = c is equal to \( m_1 = f ' (c) = 3c^2 - 10c + 7 \)
The slope \( m_2 \) of the secant through the points \( (0 , f(0)) \) and \( (3 , f(3)) \)
\[ m_2 = \dfrac{{f(3) - f(0)}}{{3 - 0}} = \dfrac{{4 - 1}}{3} = 1 \]
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 have to be equal.
\[ 3c^2 - 10c + 7 = 1 \]
which may be written as
\[ 3c^2 - 10c + 6 = 0 \]
Solve using quadratic formulas to obtain two solutions
\[ c_1 = \dfrac{{5 - \sqrt{7}}}{3} \approx 0.78 \] and \( c_2 = \dfrac{{5 + \sqrt{7}}}{3} \approx 2.55 \)
Check answer graphically
In figure 2 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 \( B(3 , f(3))\).
Figure 3. Mean Value Theorem used in example 2
