4. Applications of Derivatives

# 4.5 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 has a local extremum at a point then must be a critical point of However, a function is not guaranteed to have a local extremum at a critical point. For example, has a critical point at since is zero at but does not have a local extremum at 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 3 of the Mean Value Theorem showed that if the derivative of a function is positive over an interval then the function is increasing over On the other hand, if the derivative of the function is negative over an interval then the function is decreasing over as shown in the following figure.

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

Note that need not have a local extrema at a critical point. The critical points are candidates for local extrema only. In (Figure), we show that if a continuous function 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 has a local extremum at a critical point, then the sign of switches as increases through that point.

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

- If a continuous function has a local extremum, it must occur at a critical point
- The function has a local extremum at the critical point if and only if the derivative switches sign as increases through
- Therefore, to test whether a function has a local extremum at a critical point we must determine the sign of to the left and right of

This result is known as the **first derivative test**.

### First Derivative Test

Suppose that is a continuous function over an interval containing a critical point If is differentiable over except possibly at point then satisfies one of the following descriptions:

- If changes sign from positive when to negative when then is a local maximum of
- If changes sign from negative when to positive when then is a local minimum of
- If has the same sign for and then is neither a local maximum nor a local minimum of

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 that is continuous over an interval

- Find all critical points of and divide the interval into smaller intervals using the critical points as endpoints.
- Analyze the sign of in each of the subintervals. If is continuous over a given subinterval (which is typically the case), then the sign of in that subinterval does not change and, therefore, can be determined by choosing an arbitrary test point in that subinterval and by evaluating the sign of at that test point. Use the sign analysis to determine whether is increasing or decreasing over that interval.
- Use (Figure) and the results of step 2 to determine whether 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.

### Using the First Derivative Test to Find Local Extrema

Use the first derivative test to find the location of all local extrema for Use a graphing utility to confirm your results.

#### Solution

Step 1. The derivative is To find the critical points, we need to find where Factoring the polynomial, we conclude that the critical points must satisfy

Therefore, the critical points are Now divide the interval into the smaller intervals

Step 2. Since is a continuous function, to determine the sign of over each subinterval, it suffices to choose a point over each of the intervals and determine the sign of at each of these points. For example, let’s choose as test points.

Interval | Test Point | Sign of at Test Point | Conclusion |
---|---|---|---|

is increasing. | |||

is decreasing. | |||

is increasing. |

Step 3. Since switches sign from positive to negative as increases through has a local maximum at Since switches sign from negative to positive as increases through has a local minimum at These analytical results agree with the following graph.

Use the first derivative test to locate all local extrema for

#### Solution

has a local minimum at -2 and a local maximum at 3.

### Using the First Derivative Test

Use the first derivative test to find the location of all local extrema for Use a graphing utility to confirm your results.

#### Solution

Step 1. The derivative is

The derivative when Therefore, at The derivative is undefined at Therefore, we have three critical points: and Consequently, divide the interval into the smaller intervals and

Step 2: Since is continuous over each subinterval, it suffices to choose a test point in each of the intervals from step 1 and determine the sign of at each of these points. The points are test points for these intervals.

Interval | Test Point | Sign of at Test Point | Conclusion |
---|---|---|---|

is decreasing. | |||

is increasing. | |||

is increasing. | |||

is decreasing. |

Step 3: Since is decreasing over the interval and increasing over the interval has a local minimum at Since is increasing over the interval and the interval does not have a local extremum at Since is increasing over the interval and decreasing over the interval has a local maximum at The analytical results agree with the following graph.

Use the first derivative test to find all local extrema for

#### Solution

has no local extrema because does not change sign at

#### Hint

The only critical point of is

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

### Definition

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

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

### Test for Concavity

Let be a function that is twice differentiable over an interval

- If for all then is concave up over
- If for all then is concave down over

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

### Definition

If is continuous at and changes concavity at the point is an inflection point of

### Testing for Concavity

For the function determine all intervals where is concave up and all intervals where is concave down. List all inflection points for Use a graphing utility to confirm your results.

#### Solution

To determine concavity, we need to find the second derivative The first derivative is so the second derivative is If the function changes concavity, it occurs either when or is undefined. Since is defined for all real numbers we need only find where Solving the equation we see that is the only place where could change concavity. We now test points over the intervals and to determine the concavity of The points and are test points for these intervals.

Interval | Test Point | Sign of at Test Point | Conclusion |
---|---|---|---|

is concave down | |||

is concave up. |

We conclude that is concave down over the interval and concave up over the interval Since changes concavity at the point is an inflection point. (Figure) confirms the analytical results.

For find all intervals where is concave up and all intervals where is concave down.

#### Solution

is concave up over the interval and concave down over the interval

#### Hint

Find where

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

Sign of | Sign of | Is 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 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 be a twice-differentiable function such that and is continuous over an open interval containing Suppose Since is continuous over for all ((Figure)). Then, by Corollary 3, is a decreasing function over Since we conclude that for all if and if Therefore, by the first derivative test, has a local maximum at On the other hand, suppose there exists a point such that but Since is continuous over an open interval containing then for all ((Figure)). Then, by Corollary is an increasing function over Since we conclude that for all if and if Therefore, by the first derivative test, has a local minimum at

### Second Derivative Test

Suppose is continuous over an interval containing

- If then has a local minimum at
- If then has a local maximum at
- If then the test is inconclusive.

Note that for case iii. when then may have a local maximum, local minimum, or neither at For example, the functions and all have critical points at In each case, the second derivative is zero at However, the function has a local minimum at whereas the function has a local maximum at and the function does not have a local extremum at

Let’s now look at how to use the second derivative test to determine whether has a local maximum or local minimum at a critical point where

### Using the Second Derivative Test

Use the second derivative to find the location of all local extrema for

#### Solution

To apply the second derivative test, we first need to find critical points where The derivative is Therefore, when

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

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

Conclusion | ||
---|---|---|

Local maximum | ||

0 | 0 | Second derivative test is inconclusive |

Local minimum |

By the second derivative test, we conclude that has a local maximum at and has a local minimum at The second derivative test is inconclusive at To determine whether has a local extrema at we apply the first derivative test. To evaluate the sign of for and let and be the two test points. Since and we conclude that is decreasing on both intervals and, therefore, does not have a local extrema at as shown in the following graph.

Consider the function The points satisfy Use the second derivative test to determine whether has a local maximum or local minimum at those points.

#### Solution

has a local maximum at -2 and a local minimum at 3.

#### Hint

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

### Key Concepts

- If is a critical point of and for and for then has a local maximum at
- If is a critical point of and for and for then has a local minimum at
- If over an interval then is concave up over
- If over an interval then is concave down over
- If and then has a local minimum at
- If and then has a local maximum at
- If and then evaluate at a test point to the left of and a test point to the right of to determine whether has a local extremum at

**1.** If is a critical point of when is there no local maximum or minimum at Explain.

**2.** For the function is both an inflection point and a local maximum/minimum?

#### Solution

It is not a local maximum/minimum because does not change sign

**3.** For the function is an inflection point?

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

#### Solution

No

**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 for some value of

#### Solution

False; for example,

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

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

**8.**

#### Solution

Increasing for and decreasing for and

**9.**

**10.**

#### Solution

Decreasing for increasing for

**11.**

**12.**

#### Solution

Decreasing for and increasing for and and

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

- is increasing and decreasing and
- the minima and maxima are located.

**13.**

**14.**

#### Solution

a. Increasing over decreasing over b. maxima at and minima at and and

**15.**

**16.**

#### Solution

a. Increasing over decreasing over b. Minimum at

**17.**

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

**18.**

#### Solution

Concave up on all no inflection points

**19.**

**20.**

#### Solution

Concave up on all no inflection points

**21.**

**22.**

#### Solution

Concave up for and concave down for inflection points at and

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

**23.** over over

**24.** over over for all

#### Solution

Answers will vary

**25.** over local maximum at local minima at

**26.** There is a local maximum at local minimum at and the graph is neither concave up nor concave down.

#### Solution

Answers will vary

**27.** There are local maxima at the function is concave up for all and the function remains positive for all

For the following exercises, determine

- intervals where is increasing or decreasing and
- local minima and maxima of

**28.** over

#### Solution

a. Increasing over decreasing over b. Local maximum at local minimum at

**29.**

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

**30.**

#### Solution

a. Concave up for concave down for b. Inflection point at

For the following exercises, determine

- intervals where is increasing or decreasing,
- local minima and maxima of
- intervals where is concave up and concave down, and
- the inflection points of

**31.**

**32.**

#### Solution

a. Increasing over and decreasing over b. Maximum at minimum at c. Concave up for concave down for d. Infection point at

**33.**

**34.**

#### Solution

a. Increasing over and decreasing over b. Minimum at c. Concave down for concave up for d. Inflection point at

**35.**

**36.**

#### Solution

a. Increasing over decreasing over b. Minimum at c. Concave up for all d. No inflection points

**37.**

For the following exercises, determine

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

**38. [T]** over

#### Solution

a. Increases over decreases over and b. Minimum at maximum at c. Concave up for concave down for and d. Inflection points at

**39. [T]** over

**40. [T]** over

#### Solution

a. Increasing for all b. No local minimum or maximum c. Concave up for concave down for d. Inflection point at

**41. [T]**

**42. [T]**

#### Solution

a. Increasing for all where defined b. No local minima or maxima c. Concave up for concave down for d. No inflection points in domain

**43. [T]** over

**44.** over

#### Solution

a. Increasing over decreasing over b. Minimum at maximum at c. Concave up for concave down for d. Infection points at

**45.**

**46.**

#### Solution

a. Increasing over decreasing over b. Minimum at c. Concave up for concave down for d. Inflection point at

**47.**

For the following exercises, interpret the sentences in terms of

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

#### Solution

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

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

#### Solution

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

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

#### Solution

For the following exercises, consider a third-degree polynomial which has the properties Determine whether the following statements are *true or false*. Justify your answer.

**53.** for some

**54.** for some

#### Solution

True, by the Mean Value Theorem

**55.** There is no absolute maximum at

**56.** If has three roots, then it has 1 inflection point.

#### Solution

True, examine derivative

**57.** If has one inflection point, then it has three real roots.

## Glossary

- concave down
- if is differentiable over an interval and is decreasing over then is concave down over

- concave up
- if is differentiable over an interval and is increasing over then is concave up over

- concavity
- the upward or downward curve of the graph of a function

- concavity test
- suppose is twice differentiable over an interval if over then is concave up over if over then is concave down over

- first derivative test
- let be a continuous function over an interval containing a critical point such that is differentiable over except possibly at if changes sign from positive to negative as increases through then has a local maximum at if changes sign from negative to positive as increases through then has a local minimum at if does not change sign as increases through then does not have a local extremum at

- inflection point
- if is continuous at and changes concavity at the point is an inflection point of

- second derivative test
- suppose and is continuous over an interval containing if then has a local minimum at if then has a local maximum at if then the test is inconclusive

## Hint

Find all critical points of and determine the signs of over particular intervals determined by the critical points.