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Calculus Volume 1

4.5 Derivatives and the Shape of a Graph

Calculus Volume 14.5 Derivatives and the Shape of a Graph
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  1. Preface
  2. 1 Functions and Graphs
    1. Introduction
    2. 1.1 Review of Functions
    3. 1.2 Basic Classes of Functions
    4. 1.3 Trigonometric Functions
    5. 1.4 Inverse Functions
    6. 1.5 Exponential and Logarithmic Functions
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Chapter Review Exercises
  3. 2 Limits
    1. Introduction
    2. 2.1 A Preview of Calculus
    3. 2.2 The Limit of a Function
    4. 2.3 The Limit Laws
    5. 2.4 Continuity
    6. 2.5 The Precise Definition of a Limit
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Chapter Review Exercises
  4. 3 Derivatives
    1. Introduction
    2. 3.1 Defining the Derivative
    3. 3.2 The Derivative as a Function
    4. 3.3 Differentiation Rules
    5. 3.4 Derivatives as Rates of Change
    6. 3.5 Derivatives of Trigonometric Functions
    7. 3.6 The Chain Rule
    8. 3.7 Derivatives of Inverse Functions
    9. 3.8 Implicit Differentiation
    10. 3.9 Derivatives of Exponential and Logarithmic Functions
    11. Key Terms
    12. Key Equations
    13. Key Concepts
    14. Chapter Review Exercises
  5. 4 Applications of Derivatives
    1. Introduction
    2. 4.1 Related Rates
    3. 4.2 Linear Approximations and Differentials
    4. 4.3 Maxima and Minima
    5. 4.4 The Mean Value Theorem
    6. 4.5 Derivatives and the Shape of a Graph
    7. 4.6 Limits at Infinity and Asymptotes
    8. 4.7 Applied Optimization Problems
    9. 4.8 L’Hôpital’s Rule
    10. 4.9 Newton’s Method
    11. 4.10 Antiderivatives
    12. Key Terms
    13. Key Equations
    14. Key Concepts
    15. Chapter Review Exercises
  6. 5 Integration
    1. Introduction
    2. 5.1 Approximating Areas
    3. 5.2 The Definite Integral
    4. 5.3 The Fundamental Theorem of Calculus
    5. 5.4 Integration Formulas and the Net Change Theorem
    6. 5.5 Substitution
    7. 5.6 Integrals Involving Exponential and Logarithmic Functions
    8. 5.7 Integrals Resulting in Inverse Trigonometric Functions
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Chapter Review Exercises
  7. 6 Applications of Integration
    1. Introduction
    2. 6.1 Areas between Curves
    3. 6.2 Determining Volumes by Slicing
    4. 6.3 Volumes of Revolution: Cylindrical Shells
    5. 6.4 Arc Length of a Curve and Surface Area
    6. 6.5 Physical Applications
    7. 6.6 Moments and Centers of Mass
    8. 6.7 Integrals, Exponential Functions, and Logarithms
    9. 6.8 Exponential Growth and Decay
    10. 6.9 Calculus of the Hyperbolic Functions
    11. Key Terms
    12. Key Equations
    13. Key Concepts
    14. Chapter Review Exercises
  8. A | Table of Integrals
  9. B | Table of Derivatives
  10. C | Review of Pre-Calculus
  11. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
  12. Index

Learning Objectives

  • 4.5.1. Explain how the sign of the first derivative affects the shape of a function’s graph.
  • 4.5.2. State the first derivative test for critical points.
  • 4.5.3. Use concavity and inflection points to explain how the sign of the second derivative affects the shape of a function’s graph.
  • 4.5.4. Explain the concavity test for a function over an open interval.
  • 4.5.5. Explain the relationship between a function and its first and second derivatives.
  • 4.5.6. State the second derivative test for local extrema.

Earlier in this chapter we stated that if a function ff has a local extremum at a point c,c, then cc must be a critical point of f.f. However, a function is not guaranteed to have a local extremum at a critical point. For example, f(x)=x3f(x)=x3 has a critical point at x=0x=0 since f(x)=3x2f(x)=3x2 is zero at x=0,x=0, but ff does not have a local extremum at x=0.x=0. Using the results from the previous section, we are now able to determine whether a critical point of a function actually corresponds to a local extreme value. In this section, we also see how the second derivative provides information about the shape of a graph by describing whether the graph of a function curves upward or curves downward.

The First Derivative Test

Corollary 33 of the Mean Value Theorem showed that if the derivative of a function is positive over an interval II then the function is increasing over I.I. On the other hand, if the derivative of the function is negative over an interval I,I, then the function is decreasing over II as shown in the following figure.

This figure is broken into four figures labeled a, b, c, and d. Figure a shows a function increasing convexly from (a, f(a)) to (b, f(b)). At two points the derivative is taken and it is noted that at both f’ > 0. In other words, f is increasing. Figure b shows a function increasing concavely from (a, f(a)) to (b, f(b)). At two points the derivative is taken and it is noted that at both f’ > 0. In other words, f is increasing. Figure c shows a function decreasing concavely from (a, f(a)) to (b, f(b)). At two points the derivative is taken and it is noted that at both f’ < 0. In other words, f is decreasing. Figure d shows a function decreasing convexly from (a, f(a)) to (b, f(b)). At two points the derivative is taken and it is noted that at both f’ < 0. In other words, f is decreasing.
Figure 4.30 Both functions are increasing over the interval (a,b).(a,b). At each point x,x, the derivative f(x)>0.f(x)>0. Both functions are decreasing over the interval (a,b).(a,b). At each point x,x, the derivative f(x)<0.f(x)<0.

A continuous function ff has a local maximum at point cc if and only if ff switches from increasing to decreasing at point c.c. Similarly, ff has a local minimum at cc if and only if ff switches from decreasing to increasing at c.c. If ff is a continuous function over an interval II containing cc and differentiable over I,I, except possibly at c,c, the only way ff can switch from increasing to decreasing (or vice versa) at point cc is if ff changes sign as xx increases through c.c. If ff is differentiable at c,c, the only way that f.f. can change sign as xx increases through cc is if f(c)=0.f(c)=0. Therefore, for a function ff that is continuous over an interval II containing cc and differentiable over I,I, except possibly at c,c, the only way ff can switch from increasing to decreasing (or vice versa) is if f(c)=0f(c)=0 or f(c)f(c) is undefined. Consequently, to locate local extrema for a function f,f, we look for points cc in the domain of ff such that f(c)=0f(c)=0 or f(c)f(c) is undefined. Recall that such points are called critical points of f.f.

Note that ff need not have a local extrema at a critical point. The critical points are candidates for local extrema only. In Figure 4.31, we show that if a continuous function ff has a local extremum, it must occur at a critical point, but a function may not have a local extremum at a critical point. We show that if ff has a local extremum at a critical point, then the sign of ff switches as xx increases through that point.

A function f(x) is graphed. It starts in the second quadrant and increases to x = a, which is too sharp and hence f’(a) is undefined. In this section f’ > 0. Then, f decreases from x = a to x = b (so f’ < 0 here), before increasing at x = b. It is noted that f’(b) = 0. While increasing from x = b to x = c, f’ > 0. The function has an inversion point at c, and it is marked f’(c) = 0. The function increases some more to d (so f’ > 0), which is the global maximum. It is marked that f’(d) = 0. Then the function decreases and it is marked that f’ > 0.
Figure 4.31 The function ff has four critical points: a,b,c,andd.a,b,c,andd. The function ff has local maxima at aa and d,d, and a local minimum at b.b. The function ff does not have a local extremum at c.c. The sign of ff changes at all local extrema.

Using Figure 4.31, we summarize the main results regarding local extrema.

  • If a continuous function ff has a local extremum, it must occur at a critical point c.c.
  • The function has a local extremum at the critical point cc if and only if the derivative ff switches sign as xx increases through c.c.
  • Therefore, to test whether a function has a local extremum at a critical point c,c, we must determine the sign of f(x)f(x) to the left and right of c.c.

This result is known as the first derivative test.

Theorem 4.9

First Derivative Test

Suppose that ff is a continuous function over an interval II containing a critical point c.c. If ff is differentiable over I,I, except possibly at point c,c, then f(c)f(c) satisfies one of the following descriptions:

  1. If ff changes sign from positive when x<cx<c to negative when x>c,x>c, then f(c)f(c) is a local maximum of f.f.
  2. If ff changes sign from negative when x<cx<c to positive when x>c,x>c, then f(c)f(c) is a local minimum of f.f.
  3. If ff has the same sign for x<cx<c and x>c,x>c, then f(c)f(c) is neither a local maximum nor a local minimum of f.f.

We can summarize the first derivative test as a strategy for locating local extrema.

Problem-Solving Strategy: Using the First Derivative Test

Consider a function ff that is continuous over an interval I.I.

  1. Find all critical points of ff and divide the interval II into smaller intervals using the critical points as endpoints.
  2. Analyze the sign of ff in each of the subintervals. If ff is continuous over a given subinterval (which is typically the case), then the sign of ff in that subinterval does not change and, therefore, can be determined by choosing an arbitrary test point xx in that subinterval and by evaluating the sign of ff at that test point. Use the sign analysis to determine whether ff is increasing or decreasing over that interval.
  3. Use First Derivative Test and the results of step 22 to determine whether ff has a local maximum, a local minimum, or neither at each of the critical points.

Now let’s look at how to use this strategy to locate all local extrema for particular functions.

Example 4.17

Using the First Derivative Test to Find Local Extrema

Use the first derivative test to find the location of all local extrema for f(x)=x33x29x1.f(x)=x33x29x1. Use a graphing utility to confirm your results.

Solution

Step 1. The derivative is f(x)=3x26x9.f(x)=3x26x9. To find the critical points, we need to find where f(x)=0.f(x)=0. Factoring the polynomial, we conclude that the critical points must satisfy

3(x22x3)=3(x3)(x+1)=0.3(x22x3)=3(x3)(x+1)=0.

Therefore, the critical points are x=3,−1.x=3,−1. Now divide the interval (,)(,) into the smaller intervals (,−1),(−1,3)and(3,).(,−1),(−1,3)and(3,).

Step 2. Since ff is a continuous function, to determine the sign of f(x)f(x) over each subinterval, it suffices to choose a point over each of the intervals (,−1),(−1,3)and(3,)(,−1),(−1,3)and(3,) and determine the sign of ff at each of these points. For example, let’s choose x=−2,x=0,andx=4x=−2,x=0,andx=4 as test points.

Interval Test Point Sign of f(x)=3(x3)(x+1)f(x)=3(x3)(x+1) at Test Point Conclusion
(,−1)(,−1) x=−2x=−2 (+)()()=+(+)()()=+ ff is increasing.
(−1,3)(−1,3) x=0x=0 (+)()(+)=(+)()(+)= ff is decreasing.
(3,)(3,) x=4x=4 (+)(+)(+)=+(+)(+)(+)=+ ff is increasing.

Step 3. Since ff switches sign from positive to negative as xx increases through 1,f1,f has a local maximum at x=−1.x=−1. Since ff switches sign from negative to positive as xx increases through 3,f3,f has a local minimum at x=3.x=3. These analytical results agree with the following graph.

The function f(x) = x3 – 3x2 – 9x – 1 is graphed. It has a maximum at x = −1 and a minimum at x = 3. The function is increasing before x = −1, decreasing until x = 3, and then increasing after that.
Figure 4.32 The function ff has a maximum at x=−1x=−1 and a minimum at x=3x=3
Checkpoint 4.16

Use the first derivative test to locate all local extrema for f(x)=x3+32x2+18x.f(x)=x3+32x2+18x.

Example 4.18

Using the First Derivative Test

Use the first derivative test to find the location of all local extrema for f(x)=5x1/3x5/3.f(x)=5x1/3x5/3. Use a graphing utility to confirm your results.

Solution

Step 1. The derivative is

f(x)=53x−2/353x2/3=53x2/35x2/33=55x4/33x2/3=5(1x4/3)3x2/3.f(x)=53x−2/353x2/3=53x2/35x2/33=55x4/33x2/3=5(1x4/3)3x2/3.

The derivative f(x)=0f(x)=0 when 1x4/3=0.1x4/3=0. Therefore, f(x)=0f(x)=0 at x=±1.x=±1. The derivative f(x)f(x) is undefined at x=0.x=0. Therefore, we have three critical points: x=0,x=0, x=1,x=1, and x=−1.x=−1. Consequently, divide the interval (,)(,) into the smaller intervals (,−1),(−1,0),(0,1),(,−1),(−1,0),(0,1), and (1,).(1,).

Step 2: Since ff is continuous over each subinterval, it suffices to choose a test point xx in each of the intervals from step 11 and determine the sign of ff at each of these points. The points x=−2,x=12,x=12,andx=2x=−2,x=12,x=12,andx=2 are test points for these intervals.

Interval Test Point Sign of f(x)=5(1x4/3)3x2/3f(x)=5(1x4/3)3x2/3 at Test Point Conclusion
(,−1)(,−1) x=−2x=−2 (+)()+=(+)()+= ff is decreasing.
(−1,0)(−1,0) x=12x=12 (+)(+)+=+(+)(+)+=+ ff is increasing.
(0,1)(0,1) x=12x=12 (+)(+)+=+(+)(+)+=+ ff is increasing.
(1,)(1,) x=2x=2 (+)()+=(+)()+= ff is decreasing.

Step 3: Since ff is decreasing over the interval (,−1)(,−1) and increasing over the interval (−1,0),(−1,0), ff has a local minimum at x=−1.x=−1. Since ff is increasing over the interval (−1,0)(−1,0) and the interval (0,1),(0,1), ff does not have a local extremum at x=0.x=0. Since ff is increasing over the interval (0,1)(0,1) and decreasing over the interval (1,),f(1,),f has a local maximum at x=1.x=1. The analytical results agree with the following graph.

The function f(x) = 5x1/3 – x5/3 is graphed. It decreases to its local minimum at x = −1, increases to x = 1, and then decreases after that.
Figure 4.33 The function f has a local minimum at x=−1x=−1 and a local maximum at x=1.x=1.
Checkpoint 4.17

Use the first derivative test to find all local extrema for f(x)=x13.f(x)=x13.

Concavity and Points of Inflection

We now know how to determine where a function is increasing or decreasing. However, there is another issue to consider regarding the shape of the graph of a function. If the graph curves, does it curve upward or curve downward? This notion is called the concavity of the function.

Figure 4.34(a) shows a function ff with a graph that curves upward. As xx increases, the slope of the tangent line increases. Thus, since the derivative increases as xx increases, ff is an increasing function. We say this function ff is concave up. Figure 4.34(b) shows a function ff that curves downward. As xx increases, the slope of the tangent line decreases. Since the derivative decreases as xx increases, ff is a decreasing function. We say this function ff is concave down.

Definition

Let ff be a function that is differentiable over an open interval I.I. If ff is increasing over I,I, we say ff is concave up over I.I. If ff is decreasing over I,I, we say ff is concave down over I.I.

This figure is broken into four figures labeled a, b, c, and d. Figure a shows a function increasing convexly from (a, f(a)) to (b, f(b)). At two points the derivative is taken and both are increasing, but the one taken further to the right is increasing more. It is noted that f’ is increasing and f is concave up. Figure b shows a function increasing concavely from (a, f(a)) to (b, f(b)). At two points the derivative is taken and both are increasing, but the one taken further to the right is increasing less. It is noted that f’ is decreasing and f is concave down. Figure c shows a function decreasing concavely from (a, f(a)) to (b, f(b)). At two points the derivative is taken and both are decreasing, but the one taken further to the right is decreasing less. It is noted that f’ is increasing and f is concave up. Figure d shows a function decreasing convexly from (a, f(a)) to (b, f(b)). At two points the derivative is taken and both are decreasing, but the one taken further to the right is decreasing more. It is noted that f’ is decreasing and f is concave down.
Figure 4.34 (a), (c) Since ff is increasing over the interval (a,b),(a,b), we say ff is concave up over (a,b).(a,b). (b), (d) Since ff is decreasing over the interval (a,b),(a,b), we say ff is concave down over (a,b).(a,b).

In general, without having the graph of a function f,f, how can we determine its concavity? By definition, a function ff is concave up if ff is increasing. From Corollary 3,3, we know that if ff is a differentiable function, then ff is increasing if its derivative f(x)>0.f(x)>0. Therefore, a function ff that is twice differentiable is concave up when f(x)>0.f(x)>0. Similarly, a function ff is concave down if ff is decreasing. We know that a differentiable function ff is decreasing if its derivative f(x)<0.f(x)<0. Therefore, a twice-differentiable function ff is concave down when f(x)<0.f(x)<0. Applying this logic is known as the concavity test.

Theorem 4.10

Test for Concavity

Let ff be a function that is twice differentiable over an interval I.I.

  1. If f(x)>0f(x)>0 for all xI,xI, then ff is concave up over I.I.
  2. If f(x)<0f(x)<0 for all xI,xI, then ff is concave down over I.I.

We conclude that we can determine the concavity of a function ff by looking at the second derivative of f.f. In addition, we observe that a function ff can switch concavity (Figure 4.35). However, a continuous function can switch concavity only at a point xx if f(x)=0f(x)=0 or f(x)f(x) is undefined. Consequently, to determine the intervals where a function ff is concave up and concave down, we look for those values of xx where f(x)=0f(x)=0 or f(x)f(x) is undefined. When we have determined these points, we divide the domain of ff into smaller intervals and determine the sign of ff over each of these smaller intervals. If ff changes sign as we pass through a point x,x, then ff changes concavity. It is important to remember that a function ff may not change concavity at a point xx even if f(x)=0f(x)=0 or f(x)f(x) is undefined. If, however, ff does change concavity at a point aa and ff is continuous at a,a, we say the point (a,f(a))(a,f(a)) is an inflection point of f.f.

Definition

If ff is continuous at aa and ff changes concavity at a,a, the point (a,f(a))(a,f(a)) is an inflection point of f.f.

A sinusoidal function is shown that has been shifted into the first quadrant. The function starts decreasing, so f’ < 0 and f’’ > 0. The function reaches the local minimum and starts increasing, so f’ > 0 and f’’ > 0. It is noted that the slope is increasing for these two intervals. The function then reaches an inflection point (a, f(a)) and from here the slop is decreasing even though the function continues to increase, so f’ > 0 and f’’ < 0. The function reaches the maximum and then starts decreasing, so f’ < 0 and f’’ < 0.
Figure 4.35 Since f(x)>0f(x)>0 for x<a,x<a, the function ff is concave up over the interval (,a).(,a). Since f(x)<0f(x)<0 for x>a,x>a, the function ff is concave down over the interval (a,).(a,). The point (a,f(a))(a,f(a)) is an inflection point of f.f.

Example 4.19

Testing for Concavity

For the function f(x)=x36x2+9x+30,f(x)=x36x2+9x+30, determine all intervals where ff is concave up and all intervals where ff is concave down. List all inflection points for f.f. Use a graphing utility to confirm your results.

Solution

To determine concavity, we need to find the second derivative f(x).f(x). The first derivative is f(x)=3x212x+9,f(x)=3x212x+9, so the second derivative is f(x)=6x12.f(x)=6x12. If the function changes concavity, it occurs either when f(x)=0f(x)=0 or f(x)f(x) is undefined. Since ff is defined for all real numbers x,x, we need only find where f(x)=0.f(x)=0. Solving the equation 6x12=0,6x12=0, we see that x=2x=2 is the only place where ff could change concavity. We now test points over the intervals (,2)(,2) and (2,)(2,) to determine the concavity of f.f. The points x=0x=0 and x=3x=3 are test points for these intervals.

Interval Test Point Sign of f(x)=6x12f(x)=6x12 at Test Point Conclusion
(,2)(,2) x=0x=0 ff is concave down
(2,)(2,) x=3x=3 ++ ff is concave up.

We conclude that ff is concave down over the interval (,2)(,2) and concave up over the interval (2,).(2,). Since ff changes concavity at x=2,x=2, the point (2,f(2))=(2,32)(2,f(2))=(2,32) is an inflection point. Figure 4.36 confirms the analytical results.

The function f(x) = x3 – 6x2 + 9x + 30 is graphed. The inflection point (2, 32) is marked, and it is roughly equidistant from the two local extrema.
Figure 4.36 The given function has a point of inflection at (2,32)(2,32) where the graph changes concavity.
Checkpoint 4.18

For f(x)=x3+32x2+18x,f(x)=x3+32x2+18x, find all intervals where ff is concave up and all intervals where ff is concave down.

We now summarize, in Table 4.1, the information that the first and second derivatives of a function ff provide about the graph of f,f, and illustrate this information in Figure 4.37.

Sign of ff Sign of ff Is ff increasing or decreasing? Concavity
Positive Positive Increasing Concave up
Positive Negative Increasing Concave down
Negative Positive Decreasing Concave up
Negative Negative Decreasing Concave down
Table 4.1 What Derivatives Tell Us about Graphs
A function is graphed in the first quadrant. It is broken up into four sections, with the breaks coming at the local minimum, inflection point, and local maximum, respectively. The first section is decreasing and concave up; here, f’ < 0 and f’’ > 0. The second section is increasing and concave up; here, f’ > 0 and f’’ > 0. The third section is increasing and concave down; here, f’ > 0 and f’’ < 0. The fourth section is increasing and concave down; here, f’ < 0 and f’’ < 0.
Figure 4.37 Consider a twice-differentiable function ff over an open interval I.I. If f(x)>0f(x)>0 for all xI,xI, the function is increasing over I.I. If f(x)<0f(x)<0 for all xI,xI, the function is decreasing over I.I. If f(x)>0f(x)>0 for all xI,xI, the function is concave up. If f(x)<0f(x)<0 for all xI,xI, the function is concave down on I.I.

The Second Derivative Test

The first derivative test provides an analytical tool for finding local extrema, but the second derivative can also be used to locate extreme values. Using the second derivative can sometimes be a simpler method than using the first derivative.

We know that if a continuous function has a local extrema, it must occur at a critical point. However, a function need not have a local extrema at a critical point. Here we examine how the second derivative test can be used to determine whether a function has a local extremum at a critical point. Let ff be a twice-differentiable function such that f(a)=0f(a)=0 and ff is continuous over an open interval II containing a.a. Suppose f(a)<0.f(a)<0. Since ff is continuous over I,I, f(x)<0f(x)<0 for all xIxI (Figure 4.38). Then, by Corollary 3,3, ff is a decreasing function over I.I. Since f(a)=0,f(a)=0, we conclude that for all xI,f(x)>0xI,f(x)>0 if x<ax<a and f(x)<0f(x)<0 if x>a.x>a. Therefore, by the first derivative test, ff has a local maximum at x=a.x=a. On the other hand, suppose there exists a point bb such that f(b)=0f(b)=0 but f(b)>0.f(b)>0. Since ff is continuous over an open interval II containing b,b, then f(x)>0f(x)>0 for all xIxI (Figure 4.38). Then, by Corollary 3,f3,f is an increasing function over I.I. Since f(b)=0,f(b)=0, we conclude that for all xI,xI, f(x)<0f(x)<0 if x<bx<b and f(x)>0f(x)>0 if x>b.x>b. Therefore, by the first derivative test, ff has a local minimum at x=b.x=b.

A function f(x) is graphed in the first quadrant with a and b marked on the x-axis. The function is vaguely sinusoidal, increasing first to x = a, then decreasing to x = b, and increasing again. At (a, f(a)), the tangent is marked, and it is noted that f’(a) = 0 and f’’(a) < 0. At (b, f(b)), the tangent is marked, and it is noted f’(b) = 0 and f’’(b) > 0.
Figure 4.38 Consider a twice-differentiable function ff such that ff is continuous. Since f(a)=0f(a)=0 and f(a)<0,f(a)<0, there is an interval II containing aa such that for all xx in I,I, ff is increasing if x<ax<a and ff is decreasing if x>a.x>a. As a result, ff has a local maximum at x=a.x=a. Since f(b)=0f(b)=0 and f(b)>0,f(b)>0, there is an interval II containing bb such that for all xx in I,I, ff is decreasing if x<bx<b and ff is increasing if x>b.x>b. As a result, ff has a local minimum at x=b.x=b.
Theorem 4.11

Second Derivative Test

Suppose f(c)=0,ff(c)=0,f is continuous over an interval containing c.c.

  1. If f(c)>0,f(c)>0, then ff has a local minimum at c.c.
  2. If f(c)<0,f(c)<0, then ff has a local maximum at c.c.
  3. If f(c)=0,f(c)=0, then the test is inconclusive.

Note that for case iii. when f(c)=0,f(c)=0, then ff may have a local maximum, local minimum, or neither at c.c. For example, the functions f(x)=x3,f(x)=x3, f(x)=x4,f(x)=x4, and f(x)=x4f(x)=x4 all have critical points at x=0.x=0. In each case, the second derivative is zero at x=0.x=0. However, the function f(x)=x4f(x)=x4 has a local minimum at x=0x=0 whereas the function f(x)=x4f(x)=x4 has a local maximum at x,x, and the function f(x)=x3f(x)=x3 does not have a local extremum at x=0.x=0.

Let’s now look at how to use the second derivative test to determine whether ff has a local maximum or local minimum at a critical point cc where f(c)=0.f(c)=0.

Example 4.20

Using the Second Derivative Test

Use the second derivative to find the location of all local extrema for f(x)=x55x3.f(x)=x55x3.

Solution

To apply the second derivative test, we first need to find critical points cc where f(c)=0.f(c)=0. The derivative is f(x)=5x415x2.f(x)=5x415x2. Therefore, f(x)=5x415x2=5x2(x23)=0f(x)=5x415x2=5x2(x23)=0 when x=0,±3.x=0,±3.

To determine whether ff has a local extrema at any of these points, we need to evaluate the sign of ff at these points. The second derivative is

f(x)=20x330x=10x(2x23).f(x)=20x330x=10x(2x23).

In the following table, we evaluate the second derivative at each of the critical points and use the second derivative test to determine whether ff has a local maximum or local minimum at any of these points.

xx f(x)f(x) Conclusion
33 −303−303 Local maximum
00 00 Second derivative test is inconclusive
33 303303 Local minimum

By the second derivative test, we conclude that ff has a local maximum at x=3x=3 and ff has a local minimum at x=3.x=3. The second derivative test is inconclusive at x=0.x=0. To determine whether ff has a local extrema at x=0,x=0, we apply the first derivative test. To evaluate the sign of f(x)=5x2(x23)f(x)=5x2(x23) for x(3,0)x(3,0) and x(0,3),x(0,3), let x=−1x=−1 and x=1x=1 be the two test points. Since f(−1)<0f(−1)<0 and f(1)<0,f(1)<0, we conclude that ff is decreasing on both intervals and, therefore, ff does not have a local extrema at x=0x=0 as shown in the following graph.

The function f(x) = x5 – 5x3 is graphed. The function increases to (negative square root of 3, 10), then decreases to an inflection point at 0, continues decreasing to (square root of 3, −10), and then increases.
Figure 4.39 The function ff has a local maximum at x=3x=3 and a local minimum at x=3x=3
Checkpoint 4.19

Consider the function f(x)=x3(32)x218x.f(x)=x3(32)x218x. The points c=3,−2c=3,−2 satisfy f(c)=0.f(c)=0. Use the second derivative test to determine whether ff has a local maximum or local minimum at those points.

We have now developed the tools we need to determine where a function is increasing and decreasing, as well as acquired an understanding of the basic shape of the graph. In the next section we discuss what happens to a function as x±.x±. At that point, we have enough tools to provide accurate graphs of a large variety of functions.

Section 4.5 Exercises

194.

If cc is a critical point of f(x),f(x), when is there no local maximum or minimum at c?c? Explain.

195.

For the function y=x3,y=x3, is x=0x=0 both an inflection point and a local maximum/minimum?

196.

For the function y=x3,y=x3, is x=0x=0 an inflection point?

197.

Is it possible for a point cc to be both an inflection point and a local extrema of a twice differentiable function?

198.

Why do you need continuity for the first derivative test? Come up with an example.

199.

Explain whether a concave-down function has to cross y=0y=0 for some value of x.x.

200.

Explain whether a polynomial of degree 22 can have an inflection point.

For the following exercises, analyze the graphs of f,f, then list all intervals where ff is increasing or decreasing.

201.
The function f’(x) is graphed. The function starts negative and crosses the x axis at (−2, 0). Then it continues increasing a little before decreasing and crossing the x axis at (−1, 0). It achieves a local minimum at (1, −6) before increasing and crossing the x axis at (2, 0).
202.
The function f’(x) is graphed. The function starts negative and crosses the x axis at (−2, 0). Then it continues increasing a little before decreasing and touching the x axis at (−1, 0). It then increases a little before decreasing and crossing the x axis at the origin. The function then decreases to a local minimum before increasing, crossing the x-axis at (1, 0), and continuing to increase.
203.
The function f’(x) is graphed. The function starts negative and touches the x axis at the origin. Then it decreases a little before increasing to cross the x axis at (1, 0) and continuing to increase.
204.
The function f’(x) is graphed. The function starts positive and decreases to touch the x axis at (−1, 0). Then it increases to (0, 4.5) before decreasing to touch the x axis at (1, 0). Then the function increases.
205.
The function f’(x) is graphed. The function starts at (−2, 0), decreases to (−1.5, −1.5), increases to (−1, 0), and continues increasing before decreasing to the origin. Then the other side is symmetric: that is, the function increases and then decreases to pass through (1, 0). It continues decreasing to (1.5, −1.5), and then increase to (2, 0).

For the following exercises, analyze the graphs of f,f, then list all intervals where

  1. ff is increasing and decreasing and
  2. the minima and maxima are located.
206.
The function f’(x) is graphed. The function starts at (−2, 0), decreases for a little and then increases to (−1, 0), continues increasing before decreasing to the origin, at which point it increases.
207.
The function f’(x) is graphed. The function starts at (−2, 0), increases and then decreases to (−1, 0), decreases and then increases to an inflection point at the origin. Then the function increases and decreases to cross (1, 0). It continues decreasing and then increases to (2, 0).
208.
The function f’(x) is graphed from x = −2 to x = 2. It starts near zero at x = −2, but then increases rapidly and remains positive for the entire length of the graph.
209.
The function f’(x) is graphed. The function starts negative and crosses the x axis at the origin, which is an inflection point. Then it continues increasing.
210.
The function f’(x) is graphed. The function starts negative and crosses the x axis at (−1, 0). Then it continues increasing a little before decreasing and touching the x axis at the origin. It increases again and then decreases to (1, 0). Then it increases.

For the following exercises, analyze the graphs of f,f, then list all inflection points and intervals ff that are concave up and concave down.

211.
The function f’(x) is graphed. The function is linear and starts negative. It crosses the x axis at the origin.
212.
The function f’(x) is graphed. It is an upward-facing parabola with 0 as its local minimum.
213.
The function f’(x) is graphed. The function resembles the graph of x3: that is, it starts negative and crosses the x axis at the origin. Then it continues increasing.
214.
The function f’(x) is graphed. The function starts negative and crosses the x axis at (−0.5, 0). Then it continues increasing to (0, 1.5) before decreasing and touching the x axis at (1, 0). It then increases.
215.
The function f’(x) is graphed. The function starts negative and crosses the x axis at (−1, 0). Then it continues increasing to a local maximum at (0, 1), at which point it decreases and touches the x axis at (1, 0). It then increases.

For the following exercises, draw a graph that satisfies the given specifications for the domain x=[−3,3].x=[−3,3]. The function does not have to be continuous or differentiable.

216.

f(x)>0,f(x)>0f(x)>0,f(x)>0 over x>1,−3<x<0,f(x)=0x>1,−3<x<0,f(x)=0 over 0<x<10<x<1

217.

f(x)>0f(x)>0 over x>2,−3<x<−1,f(x)<0x>2,−3<x<−1,f(x)<0 over −1<x<2,f(x)<0−1<x<2,f(x)<0 for all xx

218.

f(x)<0f(x)<0 over −1<x<1,f(x)>0,−3<x<−1,1<x<3,−1<x<1,f(x)>0,−3<x<−1,1<x<3, local maximum at x=0,x=0, local minima at x=±2x=±2

219.

There is a local maximum at x=2,x=2, local minimum at x=1,x=1, and the graph is neither concave up nor concave down.

220.

There are local maxima at x=±1,x=±1, the function is concave up for all x,x, and the function remains positive for all x.x.

For the following exercises, determine

  1. intervals where ff is increasing or decreasing and
  2. local minima and maxima of f.f.
221.

f(x)=sinx+sin3xf(x)=sinx+sin3x over π<x<ππ<x<π

222.

f(x)=x2+cosxf(x)=x2+cosx

For the following exercises, determine a. intervals where ff is concave up or concave down, and b. the inflection points of f.f.

223.

f(x)=x34x2+x+2f(x)=x34x2+x+2

For the following exercises, determine

  1. intervals where ff is increasing or decreasing,
  2. local minima and maxima of f,f,
  3. intervals where ff is concave up and concave down, and
  4. the inflection points of f.f.
224.

f(x)=x26xf(x)=x26x

225.

f(x)=x36x2f(x)=x36x2

226.

f(x)=x46x3f(x)=x46x3

227.

f(x)=x116x10f(x)=x116x10

228.

f(x)=x+x2x3f(x)=x+x2x3

229.

f(x)=x2+x+1f(x)=x2+x+1

230.

f(x)=x3+x4f(x)=x3+x4

For the following exercises, determine

  1. intervals where ff is increasing or decreasing,
  2. local minima and maxima of f,f,
  3. intervals where ff is concave up and concave down, and
  4. the inflection points of f.f. Sketch the curve, then use a calculator to compare your answer. If you cannot determine the exact answer analytically, use a calculator.
231.

[T] f(x)=sin(πx)cos(πx)f(x)=sin(πx)cos(πx) over x=[−1,1]x=[−1,1]

232.

[T] f(x)=x+sin(2x)f(x)=x+sin(2x) over x=[π2,π2]x=[π2,π2]

233.

[T] f(x)=sinx+tanxf(x)=sinx+tanx over (π2,π2)(π2,π2)

234.

[T] f(x)=(x2)2(x4)2f(x)=(x2)2(x4)2

235.

[T] f(x)=11x,x1f(x)=11x,x1

236.

[T] f(x)=sinxxf(x)=sinxx over x=[−2π,2π]x=[−2π,2π] [2π,0)(0,2π][2π,0)(0,2π]

237.

f(x)=sin(x)exf(x)=sin(x)ex over x=[π,π]x=[π,π]

238.

f(x)=lnxx,x>0f(x)=lnxx,x>0

239.

f(x)=14x+1x,x>0f(x)=14x+1x,x>0

240.

f(x)=exx,x0f(x)=exx,x0

For the following exercises, interpret the sentences in terms of f,f,andf.f,f,andf.

241.

The population is growing more slowly. Here ff is the population.

242.

A bike accelerates faster, but a car goes faster. Here f=f= Bike’s position minus Car’s position.

243.

The airplane lands smoothly. Here ff is the plane’s altitude.

244.

Stock prices are at their peak. Here ff is the stock price.

245.

The economy is picking up speed. Here ff is a measure of the economy, such as GDP.

For the following exercises, consider a third-degree polynomial f(x),f(x), which has the properties f(1)=0,f(3)=0.f(1)=0,f(3)=0. Determine whether the following statements are true or false. Justify your answer.

246.

f(x)=0f(x)=0 for some 1x31x3

247.

f(x)=0f(x)=0 for some 1x31x3

248.

There is no absolute maximum at x=3x=3

249.

If f(x)f(x) has three roots, then it has 11 inflection point.

250.

If f(x)f(x) has one inflection point, then it has three real roots.

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