OBJECTIVES

When you finish this section, you should be able to:

1. Use Rolle’s Theorem (p. 275)
2. Work with the Mean Value Theorem (p. 276)
3. Identify where a function is increasing and decreasing (p. 279)

ORIGINS

Michel Rolle (1652–1719) was a French mathematician whose formal education was limited to elementary school. At age 24, he moved to Paris and married. To support his family, Rolle worked as an accountant, but he also began to study algebra and became interested in the theory of equations. Although Rolle is primarily remembered for the theorem proved here, he was also the first to use the notation \(\sqrt[n\kern1pt]{\,}\) for the \(n\)th root.

THEOREM Rolle’s Theorem

Let \(f\) be a function defined on a closed interval \([a,b].\) If:

• \(f\) is continuous on \([a,b]\),
• \(f\) is differentiable on \((a,b)\), and
• \(f(a)=f(b)\),

then there is at least one number \(c\) in the open interval \((a,b)\) for which \(f^\prime (c)=0\).

Proof

Because \(f\) is continuous on a closed interval \([ a,b]\), the Extreme Value Theorem guarantees that \(f\) has an absolute maximum value and an absolute minimum value on \([ a,b]\) . There are two possibilities:

1. If \(f\) is a constant function on \([ a,b]\), then \(f^\prime\)\((x) =0\) for all \(x\) in \(( a,b)\).
2. If \(f\) is not a constant function on \([a,b]\), then, because \(f(a)=f(b)\), either the absolute maximum or the absolute minimum occurs at some number \(c\) in the open interval \((a,b)\). Then \(f(c)\) is a local maximum value (or a local minimum value), and, since \(f\) is differentiable on \((a,b)\), \(f^{\prime}(c)=0\).

Using Rolle’s Theorem

Find the \(x\)-intercepts of \(f(x) = x^{2} - 5x + 6\), and show that \(f^\prime (c) = 0\) for some number \(c\) belonging to the interval formed by the two \(x\)-intercepts. Find \(c\).

Solution At the \(x\)-intercepts, \(f(x) = 0\). \[f(x)=x^{2}-5x+6=({x-2})({x-3})=0\]

Figure 23 \(f(x)=x^2-5x+6\)

So, \(x=2\) and \(x=3\) are the \(x\)-intercepts of the graph of \(f\), and \(f(2) =\) \(f(3) =0\).

Since \(f\) is a polynomial, it is continuous on the closed interval \([2,3]\) formed by the \(x\)-intercepts and is differentiable on the open interval \((2,3)\). The three conditions of Rolle’s Theorem are satisfied, guaranteeing that there is a number \(c\) in the open interval \((2,3)\) for which \(f^\prime (c) = 0\). Since \(f^\prime (x) = 2x-5,\) the number \(c\) for which \(f^\prime (x) =0\) is \(c=\dfrac{5}{2}.\)

See Figure 23 for the graph of \(f\).

NOW WORK

Problem 9.

THEOREM Mean Value Theorem

Let \(f\) be a function defined on a closed interval \([a, b]\). If:

• \(f\) is continuous on \([a, b]\)
• \(f\) is differentiable on \((a, b)\),

then there is at least one number \(c\) in the open interval \((a,b)\) for which \[\begin{equation*}{\bbox[#FAF8ED,5pt]{\bbox[5px, border:1px solid black, #F9F7ED]{ {f^\prime (c) =\dfrac{f(b) -f(a)}{b-a}}}}} \tag{1} \end{equation*}\]

Proof

We begin by finding the slope \(m_{\sec }\) of the secant line containing points \((a, f(a))\) and \((b, f(b)).\) \[ m_{\sec }=\dfrac{f(b) -f(a) }{b-a} \]

Then using the point \((a,f(a))\) and the point slope form of an equation of a line, an equation of this secant line is \[ y-f(a) =\dfrac{f(b) -f(a) }{b-a}(x-a) \]

so \[ y=f(a) +\dfrac{f(b) -f(a) }{b-a}(x-a) \]

Now we construct a function \(g\) that satisfies the conditions of Rolle’s Theorem, by defining \(g\) as \[ g(x)=f(x)-\underset{\color{#0066A7}{\hbox{Height of the secant line}\atop \hbox{containing \((a,\;f(a))\) and \((b,\;f(b))\)} }}{\underbrace{\hbox{\(\left[ {f(a)+\frac{f(b)-f(a)}{b-a}({x-a})}\right]\)} }} \]

for all \(x\) on \([ a,b]\).

Figure 25

Since \(f\) is continuous on \([a,b]\) and differentiable on \((a,b)\), it follows that \(g\) is also continuous on \([a,b]\) and differentiable on \((a,b)\). Also, \[\begin{array}{l} g\left( a \right) = f\left( a \right) - \left[ {f\left( a \right) + \frac{{f\left( b \right) - f\left( a \right)}}{{b - a}}\left( {a - a} \right)} \right] = 0 \\ g\left( b \right) = f\left( b \right) - \left[ {f\left( a \right) + \frac{{f\left( b \right) - f\left( a \right)}}{{b - a}}\left( {b - a} \right)} \right] = 0 \\ \end{array}\]

Now, since \(g\) satisfies the three conditions of Rolle’s Theorem, there is a number \(c\) in \((a,b)\) at which \(g^\prime (c)=0.\) Since \(f(a)\) and \(\dfrac{f(b)-f(a)}{b-a}\) are constants, \(g^\prime (x)\) is given by \[ g^\prime (x)=f^\prime (x)-\frac{f(b)-f(a)}{b-a} \]

Now we evaluate \(g^\prime\) at \(c\) and use the fact that \(g^\prime (c) = 0\). \[\begin{array}{l} g'\left( c \right) = f'\left( c \right) - \left[ {\frac{{f\left( b \right) - f\left( a \right)}}{{b - a}}} \right] = 0 \\ f'\left( c \right) = \frac{{f\left( b \right) - f\left( a \right)}}{{b - a}} \\ \end{array}\]

Verifying the Mean Value Theorem

Verify that the function \(f(x)=x^{3}-3x+5,\) \(-1\leq x\leq 1\) satisfies the conditions of the Mean Value Theorem. Find the number(s) \(c\) guaranteed by the Mean Value Theorem.

Figure 26 \(f(x) = {x^3} - 3x + 5,\) \(- 1 \le x \le 1\)

Solution Since \(f\) is a polynomial function, \(f\) is continuous on the closed interval \([-1,1]\) and differentiable on the open interval \((-1,1)\). The conditions of the Mean Value Theorem are met. Now, \[ f(-1) =7\qquad f(1) =3\qquad \hbox{ and }\qquad f^\prime (x) =3x^{2}-3 \]

The number(s) \(c\) in the open interval \((-1,1)\) guaranteed by the Mean Value Theorem satisfy the equation \[\begin{array}{l} \ \ \ \ f'\left( c \right) = \frac{{f\left( 1 \right) - f\left( { - 1} \right)}}{{1 - \left( { - 1} \right)}} \\ 3{c^2} - 3 = \frac{{3 - 7}}{{1 - \left( { - 1} \right)}} = \frac{{ - 4}}{2} = - 2 \\ \ \ \ \ \ \ 3{c^2} = 1 \\ \ \ \ \ \ \ \ \ \ \ c = \sqrt {\frac{1}{3}} = \frac{{\sqrt 3 }}{3}\,\,\,\,\,\,{\rm{or}}\,\,\,\,\,\,\,c = - \sqrt {\frac{1}{3}} = - \frac{{\sqrt 3 }}{3} \\ \end{array}\]

There are two numbers in the interval \((-1,1)\) that satisfy the Mean Value Theorem. See Figure 26.

NOW WORK

Problem 25.

Applying the Mean Value Theorem to Rectilinear Motion

Use the Mean Value Theorem to show that a car that travels \(110\) mi in \(2\) h must have had a speed of \(55\) miles per hour (mph) at least once during the 2 h.

Solution Let \(s=\) \(f(t)\) represent the distance \(s\) the car has traveled after \(t\) hours. Its average velocity during the time period from \(0\) to \(2\) h is \[ \hbox{Average velocity}=\frac{f(2)-f(0)}{2-0}=\frac{110-0}{2-0}=55\ {\rm mph} \]

Using the Mean Value Theorem, there is a time \(t_{0}\), \(0<t_{0}<2,\) at which \[ f^\prime (t_{0}) =\dfrac{f(2) -f( 0) }{2-0}=55 \]

That is, the car had a velocity of \(55\) mph at least once during the \(2\) hour period.

NOW WORK

Problem 49.

IN WORDS

If \(f^\prime ({x}) = 0\) for all numbers in \(({a,b}),\) then \(f\) is a constant function.

COROLLARY

If a function \(f\) is continuous on the closed interval \([a, b]\) and is differentiable on the open interval \(( a,b)\), and if \(f^\prime (x) =0\) for all numbers \(x\) in \(( a,b)\), then \(f\) is constant on \((a,b)\).

Proof

To show that \(f\) is a constant function, we must show that \(f(x_{1}) =f(x_{2})\) for any two numbers \(x_{1}\) and \(x_{2}\) in the interval \((a,b)\).

Suppose \(x_{1}<x_{2}\). The conditions of the Mean Value Theorem are satisfied on the interval \([x_{1}, x_{2}]\), so there is a number \(c\) in this interval for which \[ f^\prime ( c) =\dfrac{f( x_{2}) -f(x_{1}) }{x_{2}-x_{1}} \]

Since \(f^\prime ( c) =0\) for all numbers in \((a,b)\), we have \[\begin{array}{l} \,\,\,\,\,\,\,\,\,\,0 = \frac{{f\left( {{x_2}} \right) - f\left( {{x_1}} \right)}}{{{x_2} - {x_1}}} \\ \,\,\,\,\,\,\,\,\,\,0 = f\left( {{x_2}} \right) - f\left( {{x_1}} \right) \\ f\left( {{x_1}} \right) = f\left( {{x_2}} \right) \\ \end{array}\]

IN WORDS

If two functions \(f\) and \(g\) have the same derivative, then the functions differ by a constant, and the graph of \(f\) is a vertical shift of the graph of \(g\).

COROLLARY

If the functions \(f\) and \(g\) are differentiable on an open interval \((a,b)\) and if \(f^\prime (x) =g^\prime (x)\) for all numbers \(x\) in \((a,b)\), then there is a number \(C\) for which \(f(x) =g(x) +C\) on \((a,b)\).

Proof

We define the function \(h\) so that \(h(x) =f(x) -g(x)\). Then \(h\) is differentiable since it is the difference of two differentiable functions, and \[ h^\prime (x) =f^\prime (x) -g^\prime (x) =0 \]

Since \(h^\prime (x) =0\) for all numbers \(x\) in the interval \(( a,b)\), \(h\) is a constant function and we can write \[ h(x) =f(x) -g(x) =C \qquad \hbox{for some number }C \]

That is, \[ f(x) =g(x) +C \]

NEED TO REVIEW?

Increasing and decreasing functions are discussed in Section P.1, pp. 10-11.

COROLLARY Increasing/Decreasing Function Test

Let \(f\) be a function that is differentiable on the open interval \((a,b)\):

• If \(f^\prime (x) >0\) on \(( a,b)\), then \(f\) is increasing on \(( a,b)\).
• If \(f^\prime (x) <0\) on \(( a,b)\), then \(f\) is decreasing on \(( a,b)\).

Proof

Since \(f\) is differentiable on \(( a,b)\), it is continuous on \(( a,b)\). To show that \(f\) is increasing on \(( a,b)\), we choose two numbers \(x_{1}\) and \(x_{2}\) in \(( a,b)\), with \(x_{1}<x_{2}\). We need to show that \(f(x_{1}) <f(x_{2}).\)

Since \(x_{1}\) and \(x_{2}\) are in \(( a,b)\), \(f\) is continuous on the closed interval \([x_{1},x_{2}]\) and differentiable on the open interval \((x_{1}, x_{2})\). By the Mean Value Theorem, there is a number \(c\) in the interval \((x_{1}, x_{2})\) for which \[f^\prime\left( c \right) = \frac{{f\left( {{x_2}} \right) - f\left( {{x_1}} \right)}}{{{x_2} - {x_1}}}\]

Since \(f^\prime (x) >0\) for all \(x\) in \((a, b)\), it follows that \(f^\prime\left( c \right) > 0\). Since \(x_{1}\) < \(x_{2}\), then \(x_{2}-x_{1}>0\). As a result, \[\begin{array}{l} f\left( {{x_2}} \right) - f\left( {{x_1}} \right) > 0 \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,f\left( {{x_1}} \right) < f\left( {{x_2}} \right) \\ \end{array}\]

So, \(f\) is increasing on \((a,b)\).

NEED TO REVIEW?

Solving inequalities is discussed in Appendix A.1, pp. A-5 to A-8.

Identifying Where a Function Is Increasing and Decreasing

Determine where the function \(f(x)=2x^{3}-9x^{2}+12x-5\) is increasing and where it is decreasing.

Solution The function \(f\) is a polynomial so \(f\) is continuous and differentiable at every real number. We find \(f^\prime .\) \[ f^\prime (x) =6x^{2}-18x+12=6({x-2})({x-1}) \]

The Increasing/Decreasing Function Test states that \(f\) is increasing on intervals where \(f^\prime (x) >0\) and that \(f\) is decreasing on intervals where \(f^\prime (x) <0\). We solve these inequalities by using the numbers \(1\) and \(2\) to form three intervals, as shown in Figure 27. Then we determine the sign of \(f^\prime (x)\) on each interval, as shown in Table 1.

Figure 27

We conclude that \(f\) is increasing on the intervals \((-\infty ,1)\) and \((2,\infty)\), and \(f\) is decreasing on the interval \((1,2)\).

Identifying Where a Function Is Increasing and Decreasing

Determine where the function \(f(x)=(x^{2}-1)^{2/3}\) is increasing and where it is decreasing.

Solution \(f\) is continuous for all numbers \(x\), and \[ f^\prime (x)=\frac{2}{3}({x^{2}-1})^{-1/3}({2x})=\frac{4x}{3({x^{2}-1})^{1/3}} \]

The Increasing/Decreasing Function Test states that \(f\) is increasing on intervals where \(f^\prime (x) >0\) and decreasing on intervals where \(f^\prime (x) <0.\) We solve these inequalities by using the numbers \(-1,\) \(0,\) and \(1\) to form four intervals. Then we determine the sign of \(f^\prime\) in each interval, as shown in Table 2. We conclude that \(f\) is increasing on the intervals \((-1,0)\) and \((1,\infty)\) and that \(f\) is decreasing on the intervals \((-\infty ,-1)\) and \((0,1)\).

NOW WORK

Problem 33.

Determining Crop Yield *

A variation of the von Liebig model states that the yield \(f(x)\) of a plant, measured in bushels, responds to the amount \(x\) of potassium in a fertilizer according to the following square root model: \[ f(x) =-0.057-0.417x+0.852\sqrt{x} \]

For what amounts of potassium will the yield increase? For what amounts of potassium will the yield decrease?

Solution The yield is increasing when \(f^\prime (x)>0.\) \[ f^\prime (x) =-0.417+\dfrac{0.426}{\sqrt{x}}=\dfrac{-0.417\sqrt{x}+0.426}{\sqrt{x}} \]

Now, \(f^\prime (x) >0\) when \[\begin{array}{l} - 0.417\sqrt x + 0.426 > 0 \\ \,\,\,0.417\sqrt x - 0.426 < 0 \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\sqrt x < 1.022 \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\sqrt x < 1.044 \\ \end{array}\]

The crop yield is increasing when the amount of potassium in the fertilizer is less than \(1.044\) and is decreasing when the amount of potassium in the fertilizer is greater than \(1.044\).