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Sketch the Graph of a Continuous Function That Has Two Local Maximums but No Inflection Point

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Chapter 5.5: Derivatives and the Shape of a Graph

Learning Objectives

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

Earlier in this chapter we stated that if a function f has a local extremum at a point c, then c must be a critical point of f. However, a function is not guaranteed to have a local extremum at a critical point. For example, f(x)={x}^{3} has a critical point at x=0 since f^{\prime} (x)=3{x}^{2} is zero at x=0, but f does not have a local extremum at 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.

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)(a) shows a function f with a graph that curves upward. As x increases, the slope of the tangent line increases. Thus, since the derivative increases as x increases, {f}^{\prime } is an increasing function. We say this function f is concave up. (Figure)(b) shows a function f that curves downward. As x increases, the slope of the tangent line decreases. Since the derivative decreases as x increases, {f}^{\prime } is a decreasing function. We say this function f is concave down.

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

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

A sinusoidal function is shown that has been shifted into the first quadrant. The function starts decreasing, so 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

Figure 6. Since f\text{″}(x)>0 for x<a, the function f is concave up over the interval (\text{−}\infty ,a). Since f\text{″}(x)<0 for x>a, the function f is concave down over the interval (a,\infty ). The point (a,f(a)) is an inflection point of f.

Testing for Concavity

We now summarize, in (Figure), the information that the first and second derivatives of a function f provide about the graph of f, and illustrate this information in (Figure).

What Derivatives Tell Us about Graphs
Sign of f^{\prime} Sign of f\text{″} Is f increasing or decreasing? Concavity
Positive Positive Increasing Concave up
Positive Negative Increasing Concave down
Negative Positive Decreasing Concave up
Negative Negative Decreasing Concave down

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. 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

Figure 8. Consider a twice-differentiable function f over an open interval I. If f^{\prime} (x)>0 for all x\in I, the function is increasing over I. If f^{\prime} (x)<0 for all x\in I, the function is decreasing over I. If f\text{″}(x)>0 for all x\in I, the function is concave up. If f\text{″}(x)<0 for all x\in I, the function is concave down on 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 f be a twice-differentiable function such that {f}^{\prime }(a)=0 and f\text{″} is continuous over an open interval I containing a. Suppose f\text{″}(a)<0. Since f\text{″} is continuous over I, f\text{″}(x)<0 for all x\in I ((Figure)). Then, by Corollary 3, {f}^{\prime } is a decreasing function over I. Since {f}^{\prime }(a)=0, we conclude that for all x\in I,{f}^{\prime }(x)>0 if x<a and {f}^{\prime }(x)<0 if x>a. Therefore, by the first derivative test, f has a local maximum at x=a. On the other hand, suppose there exists a point b such that {f}^{\prime }(b)=0 but f\text{″}(b)>0. Since f\text{″} is continuous over an open interval I containing b, then f\text{″}(x)>0 for all x\in I ((Figure)). Then, by Corollary 3,{f}^{\prime } is an increasing function over I. Since {f}^{\prime }(b)=0, we conclude that for all x\in I, {f}^{\prime }(x)<0 if x<b and {f}^{\prime }(x)>0 if x>b. Therefore, by the first derivative test, f has a local minimum at 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.

Figure 9. Consider a twice-differentiable function f such that f\text{″} is continuous. Since f^{\prime} (a)=0 and f\text{″}(a)<0, there is an interval I containing a such that for all x in I, f is increasing if x<a and f is decreasing if x>a. As a result, f has a local maximum at x=a. Since f^{\prime} (b)=0 and f\text{″}(b)>0, there is an interval I containing b such that for all x in I, f is decreasing if x<b and f is increasing if x>b. As a result, f has a local minimum at x=b.

Note that for case iii. when f\text{″}(c)=0, then f may have a local maximum, local minimum, or neither at c. For example, the functions f(x)={x}^{3}, f(x)={x}^{4}, and f(x)=\text{−}{x}^{4} all have critical points at x=0. In each case, the second derivative is zero at x=0. However, the function f(x)={x}^{4} has a local minimum at x=0 whereas the function f(x)=\text{−}{x}^{4} has a local maximum at x, and the function f(x)={x}^{3} does not have a local extremum at x=0.

Let's now look at how to use the second derivative test to determine whether f has a local maximum or local minimum at a critical point c where {f}^{\prime }(c)=0.

Using the Second Derivative Test

Use the second derivative to find the location of all local extrema for f(x)={x}^{5}-5{x}^{3}.

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\to \pm \infty . At that point, we have enough tools to provide accurate graphs of a large variety of functions.

Key Concepts

2. For the function y={x}^{3}, is x=0 both an inflection point and a local maximum/minimum?

Solution

It is not a local maximum/minimum because {f}^{\prime } does not change sign

3. For the function y={x}^{3}, is x=0 an inflection point?

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

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

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

Solution

False; for example, y=\sqrt{x}.

7. Explain whether a polynomial of degree 2 can have an inflection point.

For the following exercises, analyze the graphs of {f}^{\prime }, then list all intervals where f is increasing or decreasing.

8.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).

9. 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.

10. 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.

Solution

Decreasing for x<1, increasing for x>1

11.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.

12.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}^{\prime }, then list all intervals where

  1. f is increasing and decreasing and
  2. the minima and maxima are located.

13.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.

14.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).

15.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.

16.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.

17.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}^{\prime }, then list all inflection points and intervals f that are concave up and concave down.

18.The function f'(x) is graphed. The function is linear and starts negative. It crosses the x axis at the origin.

Solution

Concave up on all x, no inflection points

19.The function f'(x) is graphed. It is an upward-facing parabola with 0 as its local minimum.

20.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.

Solution

Concave up on all x, no inflection points

21.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.

22.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=\left[-3,3\right]. The function does not have to be continuous or differentiable.

24. {f}^{\prime }(x)>0 over x>2,-3<x<-1,{f}^{\prime }(x)<0 over -1<x<2,f\text{″}(x)<0 for all x

Solution

Answers will vary

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

Solution

Answers will vary

For the following exercises, determine

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

28. f(x)= \sin x+{ \sin }^{3}x over \text{−}\pi <x<\pi

29. f(x)={x}^{2}+ \cos x

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

30. f(x)={x}^{3}-4{x}^{2}+x+2

For the following exercises, determine

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

31. f(x)={x}^{2}-6x

32. f(x)={x}^{3}-6{x}^{2}

33. f(x)={x}^{4}-6{x}^{3}

34. f(x)={x}^{11}-6{x}^{10}

35. f(x)=x+{x}^{2}-{x}^{3}

36. f(x)={x}^{2}+x+1

37. f(x)={x}^{3}+{x}^{4}

For the following exercises, determine

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

38. [T] f(x)= \sin (\pi x)- \cos (\pi x) over x=\left[-1,1\right]

39. [T] f(x)=x+ \sin (2x) over x=\left[-\frac{\pi }{2},\frac{\pi }{2}\right]

40. [T] f(x)= \sin x+ \tan x over (-\frac{\pi }{2},\frac{\pi }{2})

41. [T] f(x)={(x-2)}^{2}{(x-4)}^{2}

42. [T] f(x)=\frac{1}{1-x},x\ne 1

44. f(x)= \sin (x){e}^{x} over x=\left[\text{−}\pi ,\pi \right]

45. f(x)=\text{ln}x\sqrt{x},x>0

46. f(x)=\frac{1}{4}\sqrt{x}+\frac{1}{x},x>0

47. f(x)=\frac{{e}^{x}}{x},x\ne 0

For the following exercises, interpret the sentences in terms of f,{f}^{\prime },\text{ and }f\text{″}.

48. The population is growing more slowly. Here f is the population.

Solution

f>0,{f}^{\prime }>0,f\text{″}<0

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

50. The airplane lands smoothly. Here f is the plane's altitude.

Solution

f>0,{f}^{\prime }<0,f\text{″}<0

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

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

Solution

f>0,{f}^{\prime }>0,f\text{″}>0

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

53. f(x)=0 for some 1\le x\le 3

54. f\text{″}(x)=0 for some 1\le x\le 3

Solution

True, by the Mean Value Theorem

55. There is no absolute maximum at x=3

56. If f(x) has three roots, then it has 1 inflection point.

Solution

True, examine derivative

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

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Source: https://ecampusontario.pressbooks.pub/scccalculus1/chapter/derivatives-and-the-shape-of-a-graph/

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