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.

\[ f '(c) = \dfrac{{f(b) - f(a)}}{{b - a}} \]

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.

\( f'(c) \) 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

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.

There may be more than one value of \( x ( = c) \) that satisfies the mean value theorem, see example 2 below.

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 \( B(3 , f(3))\).

Solve Tangent Lines Problems in Calculus

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