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C. On What Interval(S) Is F Decreasing and Concave Up. Use to Justify Your Answer

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4. Applications of Derivatives

four.5 Derivatives and the Shape of a Graph

Learning Objectives

  • Explain how the sign of the first derivative affects the shape of a office'southward graph.
  • State the first derivative exam for critical points.
  • Employ concavity and inflection points to explicate how the sign of the second derivative affects the shape of a function'south graph.
  • Explain the concavity test for a function over an open interval.
  • Explain the relationship between a function and its outset and 2nd derivatives.
  • State the second derivative test for local extrema.

Earlier in this chapter we stated that if a part f has a local extremum at a point c, then c must be a critical point of f. All the same, a function is not guaranteed to have a local extremum at a critical indicate. For example, f(x)={x}^{3} has a critical signal at x=0 since f\prime (x)=3{x}^{2} is cipher at x=0, but f does not take a local extremum at x=0. Using the results from the previous section, nosotros are at present 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 up or curves downward.

The Outset Derivative Test

Corollary iii of the Hateful Value Theorem showed that if the derivative of a function is positive over an interval I then the office is increasing over I. On the other hand, if the derivative of the function is negative over an interval I, so the function is decreasing over I as shown in the post-obit effigy.

image
0. In other words, f is increasing. Figure b shows a role increasing concavely from (a, f(a)) to (b, f(b)). At ii points the derivative is taken and it is noted that at both f' > 0. In other words, f is increasing. Figure c shows a office 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." width="487" height="548"> Figure i. Both functions are increasing over the interval (a,b). At each indicate x, the derivative f\prime (x)>0. Both functions are decreasing over the interval (a,b). At each indicate x, the derivative f\prime (x)<0.

A continuous function f has a local maximum at point c if and but if f switches from increasing to decreasing at bespeak c. Similarly, f has a local minimum at c if and but if f switches from decreasing to increasing at c. If f is a continuous office over an interval I containing c and differentiable over I, except possibly at c, the only way f can switch from increasing to decreasing (or vice versa) at point c is if {f}^{\prime } changes sign every bit x increases through c. If f is differentiable at c, the just way that {f}^{\prime }. can change sign as x increases through c is if {f}^{\prime }(c)=0. Therefore, for a role f that is continuous over an interval I containing c and differentiable over I, except possibly at c, the only style f can switch from increasing to decreasing (or vice versa) is if f\prime (c)=0 or {f}^{\prime }(c) is undefined. Consequently, to locate local extrema for a function f, nosotros await for points c in the domain of f such that f\prime (c)=0 or {f}^{\prime }(c) is undefined. Recall that such points are called disquisitional points of f.

Annotation that f need non have a local extrema at a critical point. The critical points are candidates for local extrema only. In (Figure), nosotros show that if a continuous function f has a local extremum, information technology must occur at a critical point, just a part may not have a local extremum at a critical point. We show that if f has a local extremum at a disquisitional indicate, then the sign of {f}^{\prime } switches as x increases through that point.

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

This event is known as the first derivative exam.

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

Now let'south look at how to apply this strategy to locate all local extrema for particular functions.

Using the First Derivative Exam to Find Local Extrema

Use the first derivative test to notice the location of all local extrema for f(x)={x}^{3}-3{x}^{2}-9x-1. Apply a graphing utility to confirm your results.

Employ the first derivative test to locate all local extrema for f(x)=\text{−}{x}^{3}+\frac{3}{2}{x}^{2}+18x.

Solution

f has a local minimum at -2 and a local maximum at three.

Using the Starting time Derivative Exam

Use the first derivative test to find the location of all local extrema for f(x)=5{x}^{1\text{/}3}-{x}^{5\text{/}3}. Use a graphing utility to confirm your results.

Use the first derivative test to discover all local extrema for f(x)=\sqrt[3]{x-1}.

Concavity and Points of Inflection

We now know how to determine where a office is increasing or decreasing. However, there is another result to consider regarding the shape of the graph of a office. If the graph curves, does it bend upward or bend downward? This notion is chosen the concavity of the role.

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

In general, without having the graph of a office f, how tin can nosotros decide 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 part f that is twice differentiable is concave upwards when f\text{″}(x)>0. Similarly, a function f is concave down if {f}^{\prime } is decreasing. We know that a differentiable part {f}^{\prime } is decreasing if its derivative f\text{″}(x)<0. Therefore, a twice-differentiable office f is concave downward when f\text{″}(x)<0. Applying this logic is known as the concavity exam.

We conclude that we can decide the concavity of a function f past looking at the second derivative of f. In addition, we observe that a function f can switch concavity ((Effigy)). Still, a continuous function tin switch concavity only at a bespeak 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 downwards, nosotros wait for those values of x where f\text{″}(x)=0 or f\text{″}(x) is undefined. When we take determined these points, we split up the domain of f into smaller intervals and make up one's mind the sign of f\text{″} over each of these smaller intervals. If f\text{″} changes sign as we laissez passer through a signal x, then f changes concavity. It is important to remember that a function f may not change concavity at a signal x even if f\text{″}(x)=0 or f\text{″}(x) is undefined. If, however, f does alter concavity at a bespeak a and f is continuous at a, nosotros say the betoken (a,f(a)) is an inflection bespeak of f.

Testing for Concavity

We now summarize, in (Figure), the information that the first and 2nd derivatives of a role f provide about the graph of f, and illustrate this data in (Figure).

What Derivatives Tell U.s.a. about Graphs
Sign of f\prime Sign of f\text{″} Is f increasing or decreasing? Concavity
Positive Positive Increasing Concave upward
Positive Negative Increasing Concave downwards
Negative Positive Decreasing Concave up
Negative Negative Decreasing Concave down

The 2nd Derivative Exam

The showtime derivative test provides an analytical tool for finding local extrema, but the second derivative can besides be used to locate farthermost values. Using the second derivative tin can sometimes be a simpler method than using the outset derivative.

We know that if a continuous function has a local extrema, it must occur at a critical betoken. However, a part need non have a local extrema at a critical bespeak. Here nosotros examine how the 2nd derivative test can be used to determine whether a function has a local extremum at a disquisitional signal. 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 }(ten)<0 if x>a. Therefore, past the offset derivative test, f has a local maximum at x=a. On the other hand, suppose in that location exists a signal b such that {f}^{\prime }(b)=0 just f\text{″}(b)>0. Since f\text{″} is continuous over an open interval I containing b, so f\text{″}(x)>0 for all x\in I ((Figure)). Then, by Corollary 3,{f}^{\prime } is an increasing role over I. Since {f}^{\prime }(b)=0, we conclude that for all x\in I, {f}^{\prime number }(x)<0 if 10<b and {f}^{\prime }(x)>0 if x>b. Therefore, by the kickoff derivative test, f has a local minimum at x=b.

Note that for case iii. when f\text{″}(c)=0, and so f may accept 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. Withal, 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'due south at present await at how to employ the second derivative test to decide whether f has a local maximum or local minimum at a critical signal c where {f}^{\prime }(c)=0.

Using the 2nd Derivative Exam

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

We have now adult the tools we demand to determine where a function is increasing and decreasing, as well as acquired an agreement of the basic shape of the graph. In the next section we discuss what happens to a function every bit x\to \text{±}\infty . At that point, we have enough tools to provide accurate graphs of a large diverseness of functions.

Primal Concepts

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

Solution

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

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

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

5. Why do you need continuity for the first derivative examination? Come with an case.

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

Solution

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

seven. Explain whether a polynomial of degree 2 tin accept an inflection point.

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

viii.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, clarify the graphs of {f}^{\prime }, and so list all intervals where

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

thirteen.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.

sixteen.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 }, and so list all inflection points and intervals f that are concave up and concave downwardly.

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

nineteen.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 upward 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 non have to exist continuous or differentiable.

24. {f}^{\prime }(x)>0 over x>2,-3<x<-1,{f}^{\prime number }(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 upwards 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 post-obit exercises, decide a. intervals where f is concave upwards or concave down, and b. the inflection points of f.

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

For the following exercises, make up one's mind

  1. intervals where f is increasing or decreasing,
  2. local minima and maxima of f,
  3. intervals where f is concave upwards and concave downwardly, 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, make up one's mind

  1. intervals where f is increasing or decreasing,
  2. local minima and maxima of f,
  3. intervals where f is concave upwards and concave downward, and
  4. the inflection points of f. Sketch the bend, then use a calculator to compare your answer. If you cannot decide the exact answer analytically, employ 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]

xl. [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 post-obit 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 motorcar goes faster. Here f= Cycle'due south position minus Car's position.

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

Solution

f>0,{f}^{\prime }<0,f\text{″}<0

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

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

Solution

f>0,{f}^{\prime }>0,f\text{″}>0

For the following exercises, consider a 3rd-caste polynomial f(x), which has the properties {f}^{\prime }(1)=0,{f}^{\prime }(3)=0. Determine whether the following statements are truthful or imitation. 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 3 roots, so it has one inflection signal.

Solution

True, examine derivative

57. If f(x) has one inflection indicate, so information technology has three real roots.

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Source: https://opentextbc.ca/calculusv1openstax/chapter/derivatives-and-the-shape-of-a-graph/

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