When Is The Function Increasing When Given Derivative Graph

iv. Applications of Derivatives

4.five Derivatives and the Shape of a Graph

Learning Objectives

Explicate 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 2nd derivative affects the shape of a function's graph.
Explain the concavity test for a function over an open up interval.
Explain the relationship between a part and its first and 2d derivatives.
State the second derivative examination for local extrema.

Earlier in this chapter we stated that if a part $f$ has a local extremum at a betoken $c,$ then $c$ must be a critical point of $f.$ Notwithstanding, a function is non guaranteed to have a local extremum at a critical indicate. For example, $f(x)={x}^{3}$ has a disquisitional point at $x=0$ since $f\prime (x)=3{x}^{2}$ is zip at $x=0,$ merely $f$ does not have a local extremum at $x=0.$ Using the results from the previous section, we are at present able to determine whether a disquisitional point of a function actually corresponds to a local extreme value. In this section, we also see how the second derivative provides information almost the shape of a graph by describing whether the graph of a role curves upward or curves downward.

The Get-go Derivative Test

Corollary 3 of the Mean Value Theorem showed that if the derivative of a office is positive over an interval $I$ then the function is increasing over $I.$ On the other hand, if the derivative of the office is negative over an interval $I,$ then the part is decreasing over $I$ equally shown in the following effigy.

0. In other words, f is increasing. Figure b shows a function increasing concavely from (a, f(a)) to (b, f(b)). At 2 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 information technology is noted that at both f' < 0. In other words, f is decreasing." width="487" acme="548"> **Figure one.** Both functions are increasing over the interval $(a,b).$ At each point $x,$ the derivative $f\prime (x)>0.$ Both functions are decreasing over the interval $(a,b).$ At each point $x,$ the derivative $f\prime (x)<0.$

A continuous function $f$ has a local maximum at point $c$ if and only if $f$ switches from increasing to decreasing at point $c.$ Similarly, $f$ has a local minimum at $c$ if and only if $f$ switches from decreasing to increasing at $c.$ If $f$ is a continuous function over an interval $I$ containing $c$ and differentiable over $I,$ except maybe at $c,$ the simply fashion $f$ can switch from increasing to decreasing (or vice versa) at point $c$ is if ${f}^{\prime }$ changes sign as $x$ increases through $c.$ If $f$ is differentiable at $c,$ the just mode that ${f}^{\prime }.$ can modify sign equally $x$ increases through $c$ is if ${f}^{\prime }(c)=0.$ Therefore, for a function $f$ that is continuous over an interval $I$ containing $c$ and differentiable over $I,$ except possibly at $c,$ the only way $f$ tin 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 look for points $c$ in the domain of $f$ such that $f\prime (c)=0$ or ${f}^{\prime }(c)$ is undefined. Call back that such points are called disquisitional points of $f.$

Note that $f$ demand not have a local extrema at a disquisitional point. The critical points are candidates for local extrema simply. In (Effigy), we show that if a continuous function $f$ has a local extremum, it must occur at a critical point, but a function may non have a local extremum at a disquisitional point. We evidence that if $f$ has a local extremum at a critical point, then the sign of ${f}^{\prime }$ switches as $x$ increases through that point.

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

This result is known equally the first derivative test.

We tin summarize the kickoff derivative test every bit a strategy for locating local extrema.

At present let'south look at how to utilize this strategy to locate all local extrema for particular functions.

Using the Outset Derivative Examination to Find Local Extrema

Apply the outset derivative exam to detect the location of all local extrema for $f(x)={x}^{3}-3{x}^{2}-9x-1.$ Use a graphing utility to ostend your results.

Utilize 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

Employ the showtime derivative examination 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 notice all local extrema for $f(x)=\sqrt[3]{x-1}.$

Concavity and Points of Inflection

Nosotros at present know how to determine where a function is increasing or decreasing. Nonetheless, at that place is another result to consider regarding the shape of the graph of a function. If the graph curves, does it curve up or curve downwards? 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. Nosotros say this role $f$ is concave upwardly. (Figure)(b) shows a office $f$ that curves downward. Every bit $x$ increases, the gradient of the tangent line decreases. Since the derivative decreases every bit $x$ increases, ${f}^{\prime }$ is a decreasing function. We say this function $f$ is concave down.

In general, without having the graph of a part $f,$ how can we determine its concavity? Past 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. Nosotros know that a differentiable office ${f}^{\prime }$ is decreasing if its derivative $f\text{″}(10)<0.$ Therefore, a twice-differentiable part $f$ is concave down when $f\text{″}(x)<0.$ Applying this logic is known as the concavity exam.

We conclude that we can determine the concavity of a function $f$ by looking at the 2nd derivative of $f.$ In improver, nosotros observe that a function $f$ can switch concavity ((Figure)). All the same, a continuous function tin can switch concavity simply at a signal $x$ if $f\text{″}(x)=0$ or $f\text{″}(x)$ is undefined. Consequently, to make up one's mind the intervals where a part $f$ is concave upward and concave downwardly, we look for those values of $x$ where $f\text{″}(x)=0$ or $f\text{″}(x)$ is undefined. When we have determined these points, nosotros carve up 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 bespeak $x,$ then $f$ changes concavity. It is important to recall that a function $f$ may not alter concavity at a betoken $x$ even if $f\text{″}(x)=0$ or $f\text{″}(x)$ is undefined. If, however, $f$ does alter concavity at a point $a$ and $f$ is continuous at $a,$ we say the point $(a,f(a))$ is an inflection signal of $f.$

Testing for Concavity

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

What Derivatives Tell Us virtually 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

The 2nd Derivative Test

The first derivative exam provides an analytical tool for finding local extrema, merely the second derivative tin can likewise be used to locate farthermost values. Using the 2d derivative can sometimes be a simpler method than using the beginning derivative.

We know that if a continuous function has a local extrema, it must occur at a critical signal. However, a function need not take a local extrema at a disquisitional point. Hither we examine how the 2d derivative examination can be used to make up one's mind whether a role has a local extremum at a critical point. Let $f$ exist a twice-differentiable part 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 $x\in I$ ((Effigy)). Then, past 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 examination, $f$ 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 $b,$ and so $f\text{″}(x)>0$ for all $x\in I$ ((Figure)). Then, by Corollary $3,{f}^{\prime }$ is an increasing part over $I.$ Since ${f}^{\prime }(b)=0,$ we conclude that for all $x\in I,$ ${f}^{\prime }(ten)<0$ if $x<b$ and ${f}^{\prime }(x)>0$ if $x>b.$ Therefore, by the outset derivative test, $f$ has a local minimum at $x=b.$

Annotation that for case iii. when $f\text{″}(c)=0,$ then $f$ may have a local maximum, local minimum, or neither at $c.$ For case, 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 non have a local extremum at $x=0.$

Allow's now look at how to use the second derivative exam to make up one's mind whether $f$ has a local maximum or local minimum at a critical bespeak $c$ where ${f}^{\prime }(c)=0.$

Using the 2d Derivative Test

Use the 2nd derivative to find the location of all local extrema for $f(x)={x}^{5}-5{x}^{3}.$

We have now adult the tools we need to decide where a function is increasing and decreasing, as well as acquired an understanding of the basic shape of the graph. In the next department we discuss what happens to a part as $x\to \text{±}\infty .$ At that point, nosotros take 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 betoken and a local maximum/minimum?

Solution

Information technology is not a local maximum/minimum considering ${f}^{\prime }$ does not change sign

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

4. Is information technology possible for a point $c$ to be both an inflection indicate and a local extrema of a twice differentiable part?

5. Why practise yous need continuity for the first derivative examination? Come up up with an example.

vi. Explain whether a concave-downwards office has to cross $y=0$ for some value of $x.$

Solution

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

7. Explain whether a polynomial of degree 2 tin have an inflection signal.

For the following exercises, analyze the graphs of ${f}^{\prime },$ then listing 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.

x. 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 $10<1,$ 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 ${f}^{\prime },$ and so listing all intervals where

$f$ is increasing and decreasing and
the minima and maxima are located.

xiii. 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).

fifteen. 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 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

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 upwards 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 role does non have to exist continuous or differentiable.

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

Solution

Answers will vary

26. In that location is a local maximum at $x=2,$ local minimum at $x=1,$ and the graph is neither concave upward nor concave downwardly.

Solution

Answers will vary

For the post-obit exercises, determine

intervals where $f$ is increasing or decreasing and
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 upwards or concave downward, and b. the inflection points of $f.$

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

For the post-obit exercises, determine

intervals where $f$ is increasing or decreasing,
local minima and maxima of $f,$
intervals where $f$ is concave up and concave downwardly, and
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 post-obit exercises, make up one's mind

intervals where $f$ is increasing or decreasing,
local minima and maxima of $f,$
intervals where $f$ is concave upwardly and concave down, and
the inflection points of $f.$ Sketch the curve, then apply a calculator to compare your reply. If you cannot make up one's mind the exact answer analytically, utilise 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, translate 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 machine goes faster. Here $f=$ Bike's position minus Car's position.

fifty. 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 meridian. Here $f$ is the stock price.

52. The economy is picking upwardly speed. Hither $f$ is a measure of the economy, such as GDP.

Solution

$f>0,{f}^{\prime }>0,f\text{″}>0$

For the post-obit exercises, consider a 3rd-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.