OBJECTIVES

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

1. Use the First Derivative Test to find local extrema (p. 284)
2. Use the First Derivative Test with rectilinear motion (p. 286)
3. Determine the concavity of a function (p. 287)
4. Find inflection points (p. 290)
5. Use the Second Derivative Test to find local extrema (p. 291)

IN WORDS

If \(c\) is a critical number of \({f}\) and if \({f}\) is increasing to the left of \({c}\) and decreasing to the right of \({c},\) then \({f}({c})\) is a local maximum value. If \({f}\) is decreasing to the left of \({c}\) and increasing to the right of \({c}\), then \({f}({c})\) is a local minimum value.

THEOREM First Derivative Test

Let \(f\) be a function that is continuous on an interval \(I\). Suppose that \(c\) is a critical number of \(f\) and \((a,b)\) is an open interval in \(I\) containing \(c\):

• If \(f^\prime (x)>0\) for \(a<x<c\) and \(f^\prime (x) <0\) for \(c<x<b\), then \(f(c)\) is a local maximum value.
• If \(f^\prime (x)<0\) for \(a<x<c\) and \(f^\prime (x) >0\) for \(c<x<b\), then \(f(c)\) is a local minimum value.
• If \(f^\prime (x)\) has the same sign on both sides of \(c\), then \(f(c)\) is neither a local maximum value nor a local minimum value.

Partial Proof

If \(f^\prime (x)>0\) on the interval \((a,c)\), then \(f\) is increasing on \((a,c)\). Also, if \(f^\prime (x)<0\) on the interval \((c,b)\), then \(f\) is decreasing on \((c,b)\). So, for all \(x\) in \(( a,b)\), \(f(x)\leq f(c)\). That is, \(f(c)\) is a local maximum value. See Figure 31(a).

Using the First Derivative Test to Find Local Extrema

Figure 32

Find the local extrema of \(f(x)=x^{4}-4x^{3}\).

Solution Since \(f\) is a polynomial function, \(f\) is continuous and differentiable at every real number. We begin by finding the critical numbers of \(f\). \[ f^\prime (x) =4x^{3}-12x^{2}=4x^{2}({x-3}) \]

Figure 33 \(f(x)= x^4-4x^3\)

The critical numbers are \(0\) and \(3\). We use the critical numbers \(0\) and \(3\) to form three intervals, as shown in Figure 32. Then we determine where \(f\) is increasing and where it is decreasing by determining the sign of \(f^\prime (x)\) in each interval. See Table 3.

Using the First Derivative Test, \(f\) has neither a local maximum nor a local minimum at \(0\), and \(f\) has a local minimum at \(3\). The local minimum value is \(f(3) =-27\).

NOW WORK

Problem 13.

Using the First Derivative Test to Find Local Extrema

Find the local extrema of \(f(x)=x^{2/3}({x-5})\).

Solution The domain of \(f\) is all real numbers and \(f\) is continuous on its domain. \[ \begin{eqnarray*} &&\hspace{1pc} {f^\prime }(x) = x^{2/3}+\left( {\frac{2}{3}}\right) x^{-1/3}({x-5})=\frac{3x+2({x-5})}{3x^{1/3}}=\frac{5}{3}\!\left( {\frac{x-2}{x^{1/3}}}\right)\\[-10.5pt] && \color{#0066A7}{\underset{\hbox{Use the Product Rule.}}{\uparrow}} \end{eqnarray*} \]

Since \(f^\prime (2) =0\) and \(f^\prime (0)\) does not exist, the critical numbers are \(0\) and \(2\). The graph of \(f\) will have a horizontal tangent line at the point \((2,-3\sqrt[\kern-1pt3\kern1pt]{4})\) and a vertical tangent line at the point \((0,0)\).

Table 4 shows the intervals on which \(f\) is increasing and decreasing.

Figure 34 \(f(x)=x^{2/3}(x-5)\)

By the First Derivative Test, \(f\) has a local maximum at \(0\) and a local minimum at \(2\); \(f( 0) =0\) is a local maximum value and \(f(2) =-3\sqrt[3]{4}\) is a local minimum value.

NOW WORK

Problem 21.

Using the First Derivative Test with Rectilinear Motion

Suppose the distance \(s\) of an object from the origin at time \(t\geq 0,\) in seconds, is given by \[ s=t^{3}-9t^{2}+15t+3 \]

1. Determine the time intervals during which the object is moving to the right and to the left.
2. When does the object reverse direction?
3. When is the velocity of the object increasing and when is it decreasing?
4. Draw a figure that illustrates the motion of the object.
5. Draw a figure that illustrates the velocity of the object.

Solution (a) To investigate the motion, we find the velocity \(v\). \[ v =\frac{\textit{ds}}{\textit{dt}}=3t^{2}-18t+15=3(t^{2}-6t+5)=3(t-1)(t-5) \]

The critical numbers are \(1\) and \(5\). We use Table 5 to describe the motion of the object.

The object moves to the right for the first second and again after \(5 \,{\rm seconds}\). The object moves to the left on the interval \((1,5)\).

(b) The object reverses direction at \(t=1\) and \(t=5\).

(c) To determine when the velocity increases or decreases, we find the acceleration. \[ a=\frac{dv }{dt}=6t-18=6(t-3) \]

Since \(a<0\) on the interval \((0,3)\), the velocity \(v\) decreases for the first \(3 \,{\rm seconds}\). On the interval \((3,\infty), a>0\), so the velocity \(v\) increases from \(3 \,{\rm seconds}\) onward.

(d) Figure 35 illustrates the motion of the object.

(e) Figure 36 illustrates the velocity of the object.

Figure 35 \(s(t)=t^3-9t^2+15t+3\)

Figure 36 \(v(t)=3(t-1)(t-5)\)

NOW WORK

Problem 35.

spanDEFINITIONspan Concave Up; Concave Down.

Let \(f\) be a function that is continuous on a closed interval \([a,b]\) and differentiable on the open interval \((a,b)\).

• \(f\) is concave up on \(( a,b)\) if the graph of \(f\) lies above each of its tangent lines throughout \((a,b)\).
• \(f\) is concave down on \(( a,b)\) if the graph of \(f\) lies below each of its tangent lines throughout \((a,b)\).

THEOREM Test for Concavity

Let \(f\) be a function that is continuous on a closed interval \([a,b]\). Suppose \(f^\prime\) and \(f^{\prime \prime}\) exist on the open interval \((a,b)\).

• If \(f^{\prime \prime} (x)>0\) on the interval \((a,b)\), then \(f\) is concave up on \(( a,b)\).
• If \(f^{\prime \prime} (x)<0\) on the interval \((a,b)\), then \(f\) is concave down on \(( a,b)\).

Proof

Suppose \(f^{\prime \prime} (x) >0\) on the interval \(( a,b)\), and \(c\) is any fixed number in \((a,b)\). An equation of the tangent line to \(f\) at the point \((c,f(c))\) is \[ y=f(c)+f^\prime (c)(x-c) \]

We need to show that the graph of \(f\) lies above each of its tangent lines for all \(x\) in \(( a,b) .\) That is, we need to show that \[ f(x)\geq f(c)+f^\prime (c) (x-c)\qquad \hbox{ for all }x\hbox{ in }( a,b) \]

If \(x=c\), then \(f(x) = f(c)\) and we are finished.

If \(x\neq c\), then by applying the Mean Value Theorem to the function \(f\), there is a number \(x_{1}\) between \(c\) and \(x\), for which \[ f^\prime (x_{1})=\frac{f(x)-f(c)}{x-c} \]

Now we solve for \(f(x)\): \[\begin{equation*} f(x) =f( c) +f^\prime (x_{1}) (x-c) \tag{1} \end{equation*} \]

There are two possibilities: Either \(c < x_{1} < x\) or \(x < x_{1} < c\).

Suppose \(c<x_{1}<x\). Since \(f^{\prime \prime} (x) >0\) on the interval \((a,b)\), it follows that \(f^\prime\) is increasing on \((a,b)\). For \(x_{1}>c\), this means that \({f^\prime } (x_{1})>{f^\prime } (c)\). As a result, from (1), we have \[ f(x)>f(c)+f^\prime (c)(x-c) \]

That is, the graph of \(f\) lies above each of its tangent lines to the right of \(c\) in \((a,b)\).

Similarly, if \(x<x_{1}<c\), then \(f(x) >f( c)+f^\prime ( c) (x-c)\).

In all cases, \(f(x) \geq f(c) + f^\prime (c) (x-c)\) so \(f\) is concave up on \(( a,b)\).

The proof that if \(f^{\prime \prime} (x) <0,\) then \(f\) is concave down is left as an exercise. See Problem 132.

Determining the Concavity of \(f(x)=e^x\)

Figure 39 \(f(x) = e^x\) is concave up on its domain.

Show that \(f(x)=e^{x}\) is concave up on its domain.

Solution The domain of \(f(x) =e^{x}\) is all real numbers. The first and second derivatives of \(f\) are \[ f^\prime (x)=e^{x}\qquad f^{\prime \prime} (x)=e^{x} \]

Since \(f^{\prime \prime} (x) >0\) for all real numbers, by the Test for Concavity, \(f\) is concave up on its domain.

NOW WORK

Problem 45(a).

Finding Local Extrema and Determining Concavity

1. Find any local extrema of the function \(f(x)=x^{3}-6x^{2}+9x+30\).
2. Determine where \(f(x)=x^{3}-6x^{2}+9x+30\) is concave up and where it is concave down.

Solution (a) The first derivative of \(f\) is \[ f^\prime (x)=3x^{2}-12x+9=3( x-1) (x-3) \]

So, \(1\) and \(3\) are critical numbers of \(f.\) Now,

• \(f^\prime (x)=3( x-1) (x-3) >0\) if \(x<1\) or \(x>3\).
• \(f^\prime (x) <0\) if \(1<x<3\).

Figure 40 \(f(x)=x^3-6x^2+9x+30\)

So \(f\) is increasing on \(( -\infty ,1)\) and on \((3,\infty)\); \(f\) is decreasing on \((1,3)\).

At \(1,\) \(f\) has a local maximum, and at \(3\), \(f\) has a local minimum. The local maximum value is \(f( 1) = 34\); the local minimum value is \(f(3) =30\).

(b) To determine concavity, we use the second derivative: \[ f^{\prime \prime} (x)=6x-12= 6(x-2) \]

Now, we solve the inequalities \(f^{\prime \prime} (x) <0\) and \(f^{\prime \prime} (x) >0\) and use the Test for Concavity.

• \(f^{\prime \prime} (x) < 0 \qquad \hbox{ if }x<2,\qquad \hbox{ so }f \hbox{ is concave down on }( -\infty ,2)\)
• \(f^{\prime \prime} (x) > 0 \qquad \hbox{ if }x>2,\qquad \hbox{ so }f \hbox{ is concave up on }(2,\infty )\)

spanDEFINITIONspan Inflection Point

Suppose \(f\) is a function that is differentiable on an open interval \((a,b)\) containing \(c\). If the concavity of \(f\) changes at the point \((c, f(c))\), then \((c, f(c))\) is an inflection point of \(f\).

THEOREM A Condition for an Inflection Point

Let \(f\) denote a function that is differentiable on an open interval \((a,\;b)\) containing \(c\). If \((c,\;f(c))\) is an inflection point of \(f\), then either \(f^{\prime \prime} (c) =0\) or \(f^{\prime \prime}\) does not exist at \(c\).

Steps for Finding the Inflection Points of a Function \(f\)

Step 1 Find all numbers in the domain of \(f\) at which \(f^{\prime \prime} (x)=0\) or at which \(f^{\prime \prime}\) does not exist.

Step 2 Use the Test for Concavity to determine the concavity of \(f\) on both sides of each of these numbers.

Step 3 If the concavity changes, there is an inflection point; otherwise, no inflection point exists.

Finding Inflection Points

Find the inflection points of \(f(x)=x^{5/3}\).

Solution We follow the steps for finding an inflection point.

Figure 41 \(f(x)=x^{5/3}\)

Step 1 The domain of \(f\) is all real numbers. The first and second derivatives of \(f\) are \[ {f^\prime (x)=\frac{5}{3}x^{2/3}}\qquad f^{\prime \prime} (x)=\frac{10}{9}x^{-1/3}=\frac{10}{9x^{1/3}} \]

The second derivative of \(f\) does not exist when \(x =0\). So, \((0,0)\) is a possible inflection point.

Step 2 Now use the Test for Concavity.

• \(\hbox{If }x<0\quad \text{then}\quad\) \(f^{\prime\prime} (x)<0\) \(\hbox{ so }f\hbox{ is concave down on}\;(-\infty ,0).\)
• \(\hbox{If }x>0\quad \text{then}\quad\) \(f^{\prime\prime} (x)>0\) \(\hbox{ so }f\hbox{ is concave up on }\;(0,\infty ).\)

Step 3 Since the concavity of \(f\) changes at \(0\), we conclude that \((0,0)\) is an inflection point of \(f.\)

NOW WORK

Problems 45(b) and 51.

THEOREM Second Derivative Test

Let \(f\) be a function for which \(f^\prime\) and \(f^{\prime \prime}\) exist on an open interval \((a,b)\). Suppose \(c\) lies in \(( a,b)\) and is a critical number of \(f\):

• If \(f^{\prime \prime} (c) <0\), then \(f(c)\) is a local maximum value.
• If \(f^{\prime \prime} (c) >0\), then \(f(c)\) is a local minimum value.

Proof

Suppose \(f^{\prime \prime} (c) <0.\) Then \[ f^{\prime \prime} (c)=\lim\limits_{x\rightarrow c}\frac{f^\prime (x)-f^\prime (c)}{x-c}<0 \]

Now since \(\lim\limits_{x\rightarrow c}\dfrac{f^\prime (x)-f^\prime (c)}{x-c}<0\), there is an open interval about \(c\) for which \[ \frac{f^\prime (x)-f^\prime (c)}{x-c}<0 \]

everywhere in the interval, except possibly at \(c\) itself (refer to Example 7, p. 136, in Section 1.6). Since \(c\) is a critical number, \(f^\prime (c)=0\), so \[ \frac{f^\prime (x)}{x-c}<0 \]

For \(x<c\) on this interval, \(f^\prime (x)>0\), and for \(x>c\) on this interval, \(f^\prime (x)<0\). By the First Derivative Test, \(f(c)\) is a local maximum value.

Using the Second Derivative Test to Identify Local Extrema

1. Determine where \(f(x)=x-2\cos x,\) \(0\) \(\leq x\leq 2\pi\), is concave up and concave down.
2. Find any inflection points.
3. Use the Second Derivative Test to identify any local extreme values.

Solution (a) Since \(f\) is continuous on the closed interval \([0,2\pi]\) and \(f^\prime\) and \(f^{\prime \prime}\) exist on the open interval \(\left( 0,\,2\pi \right)\), we can use the Test for Concavity. The first and second derivatives of \(f\) are \[ f^\prime (x) =\dfrac{d}{\textit{dx}}(x-2\;\cos x) =1+2\;\sin x \qquad \hbox{and}\qquad f^{\prime \prime} (x) =\dfrac{d}{\textit{dx}}(1+2\;\sin\;x) =2\;\cos x \]

To determine concavity, we solve the inequalities \(f^{\prime \prime} (x) <0\) and \(f^{\prime \prime} (x) >0.\) Since \(f^{\prime \prime} (x) =2\;\cos\;x,\) we have

• \(f^{\prime \prime} (x) > 0\qquad \hbox{ when }0<x<\dfrac{\pi }{2}\quad \hbox{ and }\quad\dfrac{3\pi }{2}<x<2\pi\)
• \(f^{\prime \prime} (x) < 0\qquad \hbox{ when }\dfrac{\pi }{2}<x<\dfrac{3\pi }{2}\)

The function \(f\) is concave up on the intervals \(\left( 0,\dfrac{\pi }{2}\right)\) and \(\left( \dfrac{3\pi }{2},2\pi \right) ,\) and \(f\) is concave down on the interval \(\left( \dfrac{\pi }{2},\dfrac{3\pi }{2}\right)\).

(b) The concavity of \(f\) changes at \(\dfrac{\pi }{2}\) and \(\dfrac{3\pi }{2}\), so the points \(\left( \dfrac{\pi }{2},\dfrac{\pi }{2}\right)\) and \(\left( \dfrac{3\pi }{2},\dfrac{3\pi }{2}\right)\) are inflection points of \(f\).

(c) We find the critical numbers by solving the equation \(f^\prime(x) =0\). \[ \begin{eqnarray*} 1+2\;\sin\;x &=& 0\qquad 0\leq x\leq 2\pi \\ \sin\;x &=&-\dfrac{1}{2} \\ x=\frac{7\pi }{6}\qquad &&\hbox{ or }\qquad {x=\frac{11\pi }{6}} \end{eqnarray*} \]

Now using the Second Derivative Test, we get \[ f^{\prime \prime} \left(\dfrac{7\pi }{6}\right) =2\;\cos \left( \dfrac{7\pi }{6}\right) =- \sqrt{3}<0 \]

and \[ f^{\prime \prime} \left( \dfrac{11\pi }{6}\right) =2\;\cos \left(\dfrac{11\pi}{6}\right) = \sqrt{3}>0 \]

So, \[ f\left( \dfrac{7\pi }{6}\right) =\dfrac{7\pi }{6}-2\;\cos \dfrac{7\pi }{6}=\dfrac{7\pi }{6}+ \sqrt{3}\approx 5.4 \]

is a local maximum value, and \[ f\left( \dfrac{11\pi }{6}\right) =\dfrac{11\pi }{6}-2\cos \dfrac{11\pi }{6}=\dfrac{11\pi }{6}- \sqrt{3}\approx 4.03 \]

is a local minimum value.

NOW WORK

Problem 79.

Analyzing Monthly Sales

Unit monthly sales \(R\) of a new product over a period of time are expected to follow the logistic function \[ R=R(t) =\dfrac{20{,}000}{1+50e^{-t}}-\dfrac{20{,}000}{51}\qquad t\geq 0 \]

where \(t\) is measured in months.

1. When are the monthly sales increasing? When are they decreasing?
2. Find the rate of change of sales.
3. When is the rate of change of sales \(R^\prime\) increasing? When is it decreasing?
4. When is the rate of change of sales a maximum?
5. Find any inflection points of \(R(t)\).
6. Interpret the result found in (e).

Solution (a) We find \(R^\prime (t)\) and use the Increasing/Decreasing Function Test. \[ R^\prime (t) =\dfrac{d}{\textit{dt}}\left( \dfrac{20{,}000}{1+50e^{-t}}-\dfrac{20{,}000}{51}\right) =20{,}000\cdot \left[ \dfrac{50e^{-t}}{(1+50e^{-t}) ^{2}}\right] =\dfrac{1{,}000{,}000e^{-t}}{(1+50e^{-t}) ^{2}} \]

Since \(e^{-t}>0\) for all \(t\geq 0\), then \(R^\prime (t) >0\) for \(t\geq 0.\) The sales function \(R\) is an increasing function. So, monthly sales are always increasing.

(b) The rate of change of sales is given by the derivative \(R^\prime (t) =\dfrac{1{,}000{,}000e^{-t}}{(1+50e^{-t})^{2}},\) \(t\geq 0.\)

(c) Using the Increasing/Decreasing Function Test with \(R^\prime\), the rate of change of sales \(R^\prime\) is increasing when its derivative \(R^{\prime \prime} (t) >0;\) \(R^\prime (t)\) is decreasing when \(R^{\prime \prime} (t) <0\). \[ \begin{eqnarray*} R^{\prime \prime} (t) &=&\dfrac{d}{\textit{dt}}R^\prime (t) =1{,}000{,}000\left[ \dfrac{-e^{-t}(1+50e^{-t})^{2}+100e^{-2t}( 1+50e^{-t}) }{(1+50e^{-t})^{4}}\right] \\[12pt]&=&1{,}000{,}000e^{-t}\left[ \dfrac{-1-50e^{-t}+100e^{-t}}{(1+50e^{-t})^{3}}\right] =\dfrac{1{,}000{,}000e^{-t}}{(1+50e^{-t}) ^{3}}(50e^{-t}-1) \end{eqnarray*} \]

Since \(e^{-t}>0\) for all \(t\), the sign of \(R^{\prime \prime}\) depends on the sign of \(50e^{-t}-1\). \[ \begin{eqnarray*} \begin{array}{rl@{\qquad}rl} 50e^{-t}-1 &> 0 & 50e^{-t}-1 &<0\\[4pt] 50e^{-t} &> 1 & 50e^{-t} &<1 \\[4pt] 50 &> e^{t} & 50 &<e^{t} \\[4pt] t &< \ln 50 & t &>\ln 50 \end{array} \end{eqnarray*} \]

Since \(R^{\prime \prime} (t) >0\) for \(t<\ln 50 \approx 3.9\) and \(R^{\prime \prime} (t) <0\) for \(t>\ln 50 \approx 3.9\), the rate of change of sales is increasing for the first \(3.9\) months and is decreasing from \(3.9\) months on.

(d) The critical number of \(R^\prime\) is \(\ln 50 \approx 3.9\). Using the First Derivative Test, the rate of change of sales is a maximum about \(3.9\) months after the product is introduced.

(e) Since \(R^{\prime \prime} (t) >0\) for \(t<\ln 50\) and \(R^{\prime \prime} (t) <0\) for \(t>\ln 50\), the point \((\ln 50,9608)\) is the inflection point of \(R\).

(f) The sales function \(R\) is an increasing function, but at the inflection point \((\ln 50,9608)\) the rate of change in sales begins to decrease.

NOW WORK

Problem 105.