OBJECTIVES

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

1. Identify absolute maximum and minimum values and local extreme values of a function (p. 263)
2. Find critical numbers (p. 267)
3. Find absolute maximum and absolute minimum values (p. 268)

spanDEFINITIONspan Absolute Extrema

Let \(f\) be a function defined on an interval \(I\). If there is a number \(u\) in the interval for which \(f( u) \geq f(x)\) for all \(x\) in \(I\), then \(f\) has an absolute maximum at \(u\) and the number \(f(u)\) is called the absolute maximum value of  \(f\) on \(I\).

If there is a number \(v\) in \(I\) for which \(f( v) \leq f(x)\) for all \(x\) in \(I\), then \(f\) has an absolute minimum at \(v\) and the number \(f(v)\) is the absolute minimum value of \(f\) on \(I\).

The values \(f( u)\) and \(f( v)\) are sometimes called absolute extrema or the extreme values of \(f\) on \(I\). (“Extrema” is the plural of the Latin noun extremum.)

spanDEFINITIONspan Local Extrema

Let \(f\) be a function defined on some interval \(I\) and let \(u\) and \(v\) be numbers in \(I\). If there is an open interval in \(I\) containing \(u\) so that \(f(u)\geq f(x)\) for all \(x\) in this open interval, then \(f\) has a local maximum (or relative maximum) at \(u,\) and the number \(f( u)\) is called a local maximum value.

Similarly, if there is an open interval in \(I\) containing \(v\) so that \(f(v)\leq f(x)\) for all \(x\) in this open interval, then \(f\) has a local minimum (or a relative minimum) at \(v\), and the number \(f( v)\) is called a local minimum value.

The term local extreme value describes either a local maximum value of \(f\) or a local minimum value of \(f\).

Identifying Maximum and Minimum Values and Local Extreme Values from the Graph of a Function

Figures 10, 11, 12, 13, 14 and 15 show the graphs of six different functions. For each function:

1. Find the domain.
2. Determine where the function is continuous.
3. Identify the absolute maximum value and the absolute minimum value, if they exist.
4. Identify any local extreme values.

Solution

Figure 10 (a) Domain: \(\left[ 0,\dfrac{3\pi }{2}\right]\)
(b) Continuous on \(\left[ 0,\dfrac{3\pi }{2}\right]\)
(c) Absolute maximum value:
\(f( 0) =1\)
Absolute minimum value:
\(f( \pi ) =-1\)
(d) No local maximum value;
local minimum value: \(f( \pi ) =-1\)

Figure 11 (a) Domain: \((-1,3)\)
(b) Continuous on \(( -1,3)\)
(c) Absolute maximum value:
\(f( 0) =1\)
Absolute minimum value:
\(f(2) =-7\)
(d) Local maximum value:
\(f( 0) =1\)
Local minimum value:
\(f(2) =-7\)

Figure 12 (a) Domain: \((-1,4]\)
(b) Continuous on \((-1,4]\)
(c) No absolute maximum value;
absolute minimum value:
\(f(4) =1\)
(d) No local maximum value;
no local minimum value

Figure 13 (a) Domain: \(( 0,e)\)
(b) Continuous on \(( 0,e)\)
(c) No absolute maximum value;
no absolute minimum value
(d) No local maximum value;
no local minimum value

Figure 14 (a) Domain: \([ 0,6]\)
(b) Continuous on \([ 0,6]\) except
at \(x=2\)
(c)Absolute maximum value:
\(f(2) =6\)
Absolute minimum value:
\(f(6) =-2\)
(d) Local maximum value: \(f(2) =6\)
Local minimum value: \(f( 1) =0\)

Figure 15 (a) Domain: \([ -2,2]\)
(b) Continuous on \([ -2,2]\) except at \(x=0\)
(c) Absolute maximum value:
\(f( -2) =3\), \(f(2) =3\)
No absolute minimum value
(d) Local maximum value: \(f( 0) =2\)
No local minimum value

NOW WORK

Problem 7.

THEOREM Extreme Value Theorem

If a function \(f\) is continuous on a closed interval \([a,b]\), then \(f\) has an absolute maximum and an absolute minimum on \([a,b]\).

NOW WORK

Problem 9.

THEOREM Condition for a Local Maximum or a Local Minimum

If a function \(f\) has a local maximum or a local minimum at the number \(c\), then either \(f^\prime ( c) =0\) or \(f^\prime ( c)\) does not exist.

Proof

Suppose \(f\) has a local maximum at \(c\). Then, by definition, \[ f(c)\geq f(x) \]

for all \(x\) in some open interval containing \(c\). Equivalently, \[ f(x)-f(c)\leq 0 \]

The derivative of \(f\) at \(c\) may be written as \[ f^\prime ( c) =\lim\limits_{x\rightarrow c}\frac{f(x)-f(c)}{x-c} \]

provided the limit exists. If this limit does not exist, then \(f^\prime (c)\) does not exist and there is nothing further to prove.

If this limit does exist, then \[ \begin{equation*} \lim\limits_{x\rightarrow c^{-}}\frac{f(x)-f(c)}{x-c}={\lim\limits_{x \rightarrow c^{+}}}\frac{f(x)-f(c)}{x-c} \tag{1} \end{equation*} \]

In the limit on the left, \(x<c\) and \(f(x)-f(c)\leq 0\), so \(\dfrac{f(x)-f(c)}{ x-c}\geq 0\) and \[ \begin{equation*} \lim\limits_{x\rightarrow c^{-}}\dfrac{f(x)-f(c)}{x-c}\geq 0 \tag{2} \end{equation*} \]

[Do you see why? If the limit were negative, then there would be an open interval about \(c,\) \(x<c\), on which \(f(x) -f( c) <0;\) refer to Section 1.6, Example 7, p. 136.]

Similarly, in the on the right side of (1), \(x>c\) and \(f(x)-f(c)\leq 0\), so \(\dfrac{f(x)-f(c)}{x-c}\leq 0\) and \[ \begin{equation*} \lim\limits_{x\rightarrow c^{+}}\frac{f(x)-f(c)}{x-c}\leq 0 \tag{3} \end{equation*} \]

Since the limits (2) and (3) must be equal, we have \[ {f}^{\prime }(c)=\lim\limits_{x\rightarrow c}\frac{f(x)-f(c)}{x-c}=0 \]

The proof when \(f\) has a local minimum at \(c\) is similar and is left as an exercise (Problem 92).

THEOREM

If a differentiable function \(f\) has a local maximum or a local minimum at \(c\), then \(f^\prime ( c) =0\).

spanDEFINITIONspan Critical Number

A critical number of a function \(f\) is a number \(c\) in the domain of \(f\) for which either \(f^\prime ( c) =0\) or \( f^\prime ( c)\) does not exist.

Finding Critical Numbers

Find any critical numbers of the following functions:

1. \(f(x)=x^{3}-6x^{2}+9x+2\)
2. \(R(x)=\dfrac{1}{x-2}\)
3. \(g(x)=\dfrac{(x-2)^{2/3}}{x}\)
4. \(G(x)={\sin }\,x\)

Solution (a) Since \(f\) is a polynomial, it is differentiable at every real number. So, the critical numbers occur where \(f^\prime (x) =0\). \[ f^\prime (x) ={3x}^{2}-12x+9=3({x-1})({x-3}) \]

\(f^\prime (x)=0\) at \(x=1\) and \(x=3\); the numbers \(1\) and \(3\) are the critical numbers of \(f\).

(b) The domain of \(R(x) = \dfrac{1}{x-2}\) is \(\{ x\,|\,x\neq 2\} \), and \(R^\prime (x)=-\dfrac{\,1}{( x-2) ^{2} }\). \(\ R^\prime \) exists for all numbers \(x\) in the domain of \(R\) (remember, \( 2 \) is not in the domain of \(R\)). Since \(R^\prime\) is never \(0\), \(R\) has no critical numbers.

(c) The domain of \(g(x)=\dfrac{(x\,{-}\,2)^{2/3}}{x}\) is \(\{ x\,|\,x\neq 0\} ,\) and the derivative of \(g\) is \[ \begin{eqnarray*} {{g^\prime }(x)}={\frac{x\cdot \left[ {\dfrac{2}{3}}({x-2})^{-1/3}\right] -1\cdot ({x-2})^{2/3}}{x^{2}}} \underset{\underset{{\color{#0066A7}{\hbox{Multiply by} \dfrac{3(x-2) ^{1/3}}{3( x-2) ^{1/3}}}}}{\color{#0066A7}{\uparrow}}}{=} {\frac{2x-3({x-2})}{3x^{2}({x-2})^{1/3}}=\frac{6-x}{3x^{2}({x-2})^{1/3}}}\\ \end{eqnarray*} \]

Critical numbers occur where \(g^\prime (x)\) \(=0\) or where \( g^\prime (x)\) does not exist. If \(x=6\), then \(g^\prime (6) =0\). Next, \(g^\prime (x)\) does not exist where \[ \begin{eqnarray*} \begin{array}{rl@{\qquad}r@{\qquad}rr} 3x^{2}( x-2) ^{1/3} &= 0 \\ 3x^{2} &= 0 & \hbox{ or } & ( x-2) ^{1/3} &= 0 \\ x &= 0 & \hbox{ or } & x &= 2 \end{array} \end{eqnarray*} \]

We ignore \(0\) since it is not in the domain of \(g\). The critical numbers of \( g\) are \(6\) and \(2\).

(d) The domain of \(G\) is all real numbers. The function \(G\) is differentiable on its domain, so the critical numbers occur where \(G^\prime (x) =0\). \(G^\prime (x)=\cos \,x\) and \(\cos x=0\) at \(x=\pm \dfrac{\pi }{2}\), \({\pm }\dfrac{3\pi }{2}\), \({\pm }\dfrac{5\pi }{2},\ldots. \) This function has infinitely many critical numbers.

NOW WORK

Problem 25.

THEOREM Locating Extreme Values

Let \(f\) be a function that is continuous on a closed interval \( [a,b]\). Then the absolute maximum value and the absolute minimum value of \(f\) are the largest and the smallest values, respectively, found among the following:

• The values of \(f\) at the critical numbers in the open interval \((a,b)\)
• \(f(a)\) and \(f(b)\), the values of \(f\) at the endpoints \(a\) and \(b\)

Steps for Finding the Absolute Extreme Values of a Function \( f\) That Is Continuous on a Closed Interval \({[ a,b]} \)

Step 1 Locate all critical numbers in the open interval \(( a,b) \).

Step 2 Evaluate \(f\) at each critical number and at the endpoints \(a\) and \(b\).

Step 3 The largest value is the absolute maximum value; the smallest value is the absolute minimum value.

Finding Absolute Maximum and Minimum Values

Find the absolute maximum value and the absolute minimum value of each function:

1. \(f(x)=x^{3}-6x^{2}+9x+2\) on \([0,2]\)
2. \(g(x)= \dfrac{(x-2)^{2/3}}{x}\) on \([1,10]\)

Solution (a) The function \(f\), a polynomial function, is continuous on the closed interval \([0,2]\), so the Extreme Value Theorem guarantees that \(f\) has an absolute maximum value and an absolute minimum value on the interval. We follow the steps for finding the absolute extreme values to identify them.

Step 1 From Example 2(a), the critical numbers of \(f\) are \(1\) and \(3\). We exclude \(3\), since it is not in the interval \((0,2) \).

Step 2 Find the value of \(f\) at the critical number \(1\) and at the endpoints \(0\) and \(2\): \[ f(1)=6\qquad f( 0) =2\qquad f(2) =4 \]

Step 3 The largest value \(6\) is the absolute maximum value of \(f\); the smallest value \(2\) is the absolute minimum value of \(f\).

(b) The function \(g\) is continuous on the closed interval \([1,10]\), so \(g\) has an absolute maximum and an absolute minimum on the interval.

Step 1 From Example 2(c), the critical numbers of \(g\) are \(2\) and \(6\). Both critical numbers are in the interval \((1,10)\).

Step 2 We evaluate \(g\) at the critical numbers \(2\) and \(6\) and at the endpoints \(1\) and \(10\):

Step 3 The largest value \(1\) is the absolute maximum value; the smallest value \(0\) is the absolute minimum value.

NOW WORK

Problem 49.

Finding Absolute Maximum and Minimum Values

Find the absolute maximum value and absolute minimum value of the function \[ f(x)=\left\{ {{ \begin{array}{@{}l@{\qquad}l@{\quad}l} {2x-1} & \text{if} & {0\leq x\leq 2} \\ {x^{2}-5x+9} & \text{if} & {2<x\leq 3} \end{array} }}\right. \]

Solution The function \(f\) is continuous on the closed interval \([0,3]\). (You should verify this.) To find the absolute maximum value and absolute minimum value, we follow the three-step procedure.

Step 1 Find the critical numbers in the open interval \((0,3)\):

• On the open interval \((0, 2)\): \(f(x)= 2x-1\) and \(f^\prime (x) =2\). Since \(f^\prime (x) \neq 0\) on the interval \((0, 2) ,\) there are no critical numbers in \((0, 2)\).
• On the open interval \((2, 3)\): \(f(x)=x^{2}-5x+9\) and \(f^\prime (x) =2x - 5\). Solving \(f^\prime (x) =2x-5=0\), we find \(x=\dfrac{5}{2}\). Since \(\dfrac{5}{2}\) is in the interval \((2, 3)\), \(\dfrac{5}{2}\) is a critical number.
• At \(x=2\), the rule for \(f\) changes, so we investigate the one-sided limits of \(\dfrac{f(x) -f(2) }{x-2}.\) \[ \begin{eqnarray*} \lim\limits_{x\rightarrow 2^{-}}\dfrac{f(x) -f(2) }{ x-2} &=& \lim\limits_{x\rightarrow 2^{-}}\dfrac{(2x-1) -3}{x-2} =\lim\limits_{x\rightarrow 2^{-}}\dfrac{2x-4}{x-2}\\[3pt] &=&\lim\limits_{x\rightarrow 2^{-}}\dfrac{2(x-2) }{x-2}=2 \\[3pt] \lim\limits_{x\rightarrow 2^{+}}\dfrac{f(x) -f(2) }{ x-2} &=& \lim\limits_{x\rightarrow 2^{+}}\dfrac{(x^{2}-5x+9) -3}{x-2} =\lim\limits_{x\rightarrow 2^{+}}\dfrac{x^{2}-5x+6}{x-2}\\[3pt] &=&\lim\limits_{x \rightarrow 2^{+}}\dfrac{( x-2) (x-3) }{x-2} =\lim\limits_{x\rightarrow 2^{+}}(x-3) =-1 \end{eqnarray*} \]

The one-sided limits are not equal, so the derivative does not exist at \(2;\) \(2\) is a critical number.

Figure 18 \[ \begin{equation*} f(x) =\left\{\begin{array}{lll} 2x-1&{\rm if}&0\leq x\leq 2\\[4pt] x^{2}-5x+9&{\rm if}&2 < x \leq 3 \end{array}\right. \end{equation*} \]

Step 2 Evaluate \(f\) at the critical numbers \(\dfrac{5}{2}\) and \(2\) and at the endpoints \(0\) and \(3\).

Step 3 The largest value \(3\) is the absolute maximum value; the smallest value \(-1\) is the absolute minimum value.

NOW WORK

Problem 61.

Constructing a Rain Gutter

Figure 19

A rain gutter is to be constructed using a piece of aluminum \(12\) in wide. After marking a length of \(4 \,{\rm in}\). from each edge, the piece of aluminum is bent up at an angle \(\theta\), as illustrated in Figure 19. The area \(A\) of a cross section of the opening, expressed as a function of \(\theta ,\) is \[ A(\theta) =16~\sin ~\theta ( \cos \theta +1)\qquad 0 \leq \theta \leq \dfrac{\pi }{2} \]

Find the angle \(\theta\) that maximizes the area \(A\). (This bend will allow the most water to flow through the gutter.)

Solution The function \(A=A(\theta ) \) is continuous on the closed interval \(\left[ 0,\dfrac{\pi }{2}\right] \). To find the angle \(\theta\) that maximizes \(A\), we follow the three-step procedure.

Step 1 We locate all critical numbers in the open interval \(\left( 0,\dfrac{\pi }{2}\right)\). \[ \begin{eqnarray*} A^\prime (\theta ) &=& 16\sin \theta (-\!\sin \theta) +16\cos \theta ( \cos \theta +1) \hspace{12pc} \color{#0066A7}{{\hbox{Product Rule}}}\\[3pt] &=& 16[-\!\sin ^{2}\theta +\cos^{2}\theta +\cos \theta] \\[3pt] &=& 16[ ( \cos ^{2}\theta -1) +\cos ^{2}\theta +\cos \theta ] \hspace{15pc} \color{#0066A7}{{\hbox{$\kern-.5pt-\!\sin^2\theta=\cos^2\theta-1$}}}\\[3pt] &=& 16[2\cos^{2}\theta +\cos \theta -1] = 16(2\cos \theta -1) (\cos \theta +1) \end{eqnarray*} \]

The critical numbers satisfy the equation \(A^\prime (\theta ) =0, 0<\theta<\frac{\pi}{2}\). \[ \begin{eqnarray*} \begin{array}{rl@{ }l@{ }rl} 16(2\;\cos \theta -1) ( \cos \theta +1) &= 0 \\[3pt] 2\;\cos \theta -1 &= 0 & \quad\hbox{ or }\quad & \cos \theta +1 &=0 \\[3pt] \cos \theta &= \dfrac{1}{2} & \hbox{ }& \cos \theta &=-1\\[3pt] \theta &= \dfrac{\pi }{3} \;{\rm or}\; \dfrac{5\pi }{3} & \hbox{ }& \theta &= \pi \end{array} \end{eqnarray*} \]

Of these solutions, only \(\dfrac{\pi }{3}\) is in the interval \(\left( 0,\dfrac{\pi }{2}\right).\) So, \(\dfrac{\pi }{3}\) is the only critical number.

Step 2 We evaluate \(A\) at the critical number \(\dfrac{\pi }{3}\) and at the endpoints \(0\) and \(\dfrac{\pi }{2}\).

Step 3 If the aluminum is bent at an angle of \(\dfrac{\pi }{3}\), the area of the opening is maximum. The maximum area is about \(20.8 \,{\rm in}^{2}\).

NOW WORK

Problem 71.

Analyzing a Cough

Coughing is caused by increased pressure in the lungs and is accompanied by a decrease in the diameter of the windpipe. See Figure 20. The radius \(r\) of the windpipe decreases with increased pressure \(p\) according to the formula \(r_{0}-r=cp,\) where \(r_{0}\) is the radius of the windpipe when there is no difference in pressure and \(c\) is a positive constant. The volume \(V\) of air flowing through the windpipe is \[ V=kpr^{4} \]

where \(k\) is a constant. Find the radius \(r\) that allows the most air to flow through the windpipe. Restrict \(r\) so that \(0<\dfrac{r_{0}}{2}\leq r\leq r_{0}.\)

Solution

Since \(p=\) \(\dfrac{r_{0}-r}{c},\) we can express \(V\) as a function of \(r\): \[ V=V(r) =k\!\left( \dfrac{r_{0}-r}{c}\right) r^{4}=\dfrac{k r_{0}}{c} r^{4}-\dfrac{k}{c} r^{5} \qquad \dfrac{r_{0}}{2}\leq r\leq r_{0} \]

Now we find the absolute maximum of \(V\) on the interval \(\left[\dfrac{r_{0}}{2},r_{0}\right].\) \[ V^\prime (r) =\dfrac{4k\,r_{0}}{c}\,r^{3}-\dfrac{5k}{c}\,r^{4}=\dfrac{k}{c}\,r^{3}(4r_{0}-5r) \]

The only critical number in the interval \(\left( \dfrac{r_{0}}{2}, r_{0}\right)\) is \(r=\dfrac{4r_{0}}{5}.\)

We evaluate \(V\) at the critical number and at the endpoints, \(\dfrac{r_{0}}{2}\) and \(r_{0}.\)

The largest of these three values is \(\dfrac{256\,kr_{0}^{5}}{3125c}.\) So, the maximum air flow occurs when the radius of the windpipe is \(\dfrac{4r_{0}}{5},\) that is, when the windpipe contracts by \(20\%.\)