1 / 33

# Chapter 5 Graphing and Optimization - PowerPoint PPT Presentation

Chapter 5 Graphing and Optimization. Section 2 Second Derivative and Graphs. Objectives for Section 5.2 Second Derivatives and Graphs. The student will be able to use concavity as a graphing tool. The student will be able to find inflection points.

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## PowerPoint Slideshow about 'Chapter 5 Graphing and Optimization' - flavia-miranda

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

### Chapter 5Graphing and Optimization

Section 2

Second Derivativeand Graphs

Objectives for Section 5.2 Second Derivatives and Graphs

• The student will be able to use concavity as a graphing tool.

• The student will be able to find inflection points.

• The student will be able to analyze graphs and do curve sketching.

• The student will be able to find the point of diminishing returns.

The term concave upward (or simply concave up) is used to describe a portion of a graph that opens upward. Concave down(ward) is used to describe a portion of a graph that opens downward.

Concave down

Concave up

A graph is concave up on the interval (a,b) if any secant connecting two points on the graph in that interval lies above the graph.

It is concave down on (a,b) if all secants lie below the graph.

down

up

The graph of a function f is concave upward on the interval (a,b) if f´(x) is increasing on (a, b), and is concave downward on the interval (a, b) if f´(x) is decreasing on (a, b).

For y = f (x), the second derivative of f, provided it exists, is the derivative of the first derivative:

The graph of a function f is concave upward on the interval (a, b) if f´´(x) is positive on (a, b), and is concave downward on the interval (a, b) if f´´(x) is negative on (a,b).

Find the intervals where the graph of

f(x) = x3 + 24x2 + 15x – 12.

is concave up or concave down.

Find the intervals where the graph of

f(x) = x3 + 24x2 + 15x – 12.

is concave up or concave down.

f ´(x) = 3x2 + 48x + 15

f ´´(x) = 6x + 48

f ´´(x) is positive when 6x + 48 > 0 or x > –8, so it is concave up on the region (–8, ∞).

f ´´(x) is negative when 6x + 48 < 0 or x < –8, so it is concave down on the region (–∞, –8).

Example 1(continued)

f (x)

f ´´(x)

- 8

–25 < x < 20,– 400 < y <14,000

–10 < x < 1–2 < y < 6

An inflection point is a point on the graph where the concavity changes from upward to downward or downward to upward.

This means that if f´´(x) exists in a neighborhood of an inflection point, then it must change sign at that point.

Theorem 1. If y = f (x) is continuous on (a, b) and has an inflection point at x = c, then either f´´(c) = 0 or f´´(c) does not exist.

continued

Inflection Points(continued)

The theorem means that an inflection point can occur only at critical value of f´´. However, not every critical value produces an inflection point.

A critical value c for f´´produces an inflection point for the graph of f only if f´´ changes sign at c, and c is in the domain of f.

Assume that f satisfies one of the conditions in the table below, for all x in some interval (a,b). Then the other condition(s) to the right of it also hold.

Find the inflection points of f (x) = x3 + 24x2 + 15x – 12.

Find the inflection points of f (x) = x3 + 24x2 + 15x – 12.

Solution:

In example 1, we saw that f ´´(x) was negative to the left of –8 and positive to the right of –8. At x = – 8, f´´(x) = 0.

This is an inflection point because f changes from concave down to concave up at this point.

Example 2(continued)

Find the inflection point using a graphing calculator.

Inflection points can be difficult to recognize on a graphing calculator, but they are easily located using root approximation routines. For instance, the above example when f is graphed shows an inflection point somewhere between –6 and –10.

–25 < x < 20,– 400 < y <14,000

f (x)

continued

Example 2(continued)

Graphing the second derivative and using the zeros command on the calc menu shows the inflection point at –8 quite easily, because inflection points occur where the second derivative is zero.

–10 < x < 1– 2 < y < 6

f ´´(x) = 6x + 48

-8

Let c be a critical value for f (x), then

Graphing calculators and computers produce the graph of a function by plotting many points. Although quite accurate, important points on a plot may be difficult to identify. Using information gained from the function and its dervative, we can sketch by hand a very good representation of the graph of f (x). This process is called curve sketching and is summarized on the following slides.

• Step 1. Analyze f (x). Find the domain and the intercepts. The x intercepts are the solutions to f (x) = 0, and the y intercept is f (0).

• Step 2. Analyze f ´(x). Find the partition points and critical values of f ´(x). Construct a sign chart for f ´(x), determine the intervals where f is increasing and decreasing, and find local maxima and minima.

Graphing Strategy(continued)

• Step 3. Analyze f´´(x). Find the partition numbers of f´´(x). Construct a sign chart for f´´(x), determine the intervals where the graph of f is concave upward and concave downward, and find inflection points.

• Step 4. Sketch the graph of f. Locate intercepts, local maxima and minima, and inflection points. Sketch in what you know from steps 1-3. Plot additional points as needed and complete the sketch.

Graphing StrategyExample

• Sketch the graph of y = x3/3 – x2 – 3x

• Step 1. Analyze f (x). This is a polynomial function, so the domain is all reals. The y intercept is 0, and the x intercepts are 0 and

• Step 2. Analyze f´(x). f ´(x) = x2 – 2x – 3 = (x+1)(x–3), so f has critical values at –1 and 3.

• Step 3. Analyze f´´(x). f ´´(x) = 2x – 2, so f´´ has a critical value at x = 1.

A combined (steps 2 and 3) sign chart for this function is shown on the next slide.

Sign chart for f´ and f´´

(– ∞, –1) (–1, 3) (3, ∞)

f ´´(x) - - - - - - - 0 + + + + + + + +

f ´(x) + + + 0 - - - - - - 0 + + + + +

- 1 1 3

f (x) increasing decreasing increasing

f (x) maximum minimum

f (x) concave down - inflection - concave up

point

A company estimates that it will sell N(x) units of a product after spending \$x thousand on advertising, as given by

N(x) = –2x3 + 90x2 – 750x + 2000 for 5 ≤ x ≤ 25

(a) When is the rate of change of sales, N ´(x), increasing? Decreasing?

A company estimates that it will sell N(x) units of a product after spending \$x thousand on advertising, as given by

N(x) = –2x3 + 90x2 – 750x + 2000 for 5 ≤ x ≤ 25

(a) When is the rate of change of sales, N ´(x), increasing? Decreasing?

–5 < x < 50 and –1000 < y < 1000

N ´(x) = –6x2 + 180x –750.

N ´(x) is increasing on (5, 15), then decreases for (15, 25).

15

Note: This is the graph of the derivative of N ´(x)

Application(continued)

• Find the inflection points for the graph of N.

0 < x < 70 and –0.03 < y < 0.015

15

15

Note: This is N (x).

Note: This is N ´(x).

Application(continued)

• Find the inflection points for the graph of N.

• N ´(x) = –6x2 + 180x –750.

• N ´´(x) = –12x + 180

• Inflection point at x = 15.

Application(continued)

(c) What is the maximum rate of change of sales?

Application(continued)

(c) What is the maximum rate of change of sales?

We want the maximum of the derivative.

N ´(x) = –6x2 + 180x –750.

Maximum at x = 15.

N ´(15) = 600.

- 5 < x < 50 and – 1000 < y < 1000

Note: This is the graph of N ´(x).

If a company decides to increase spending on advertising, they would expect sales to increase. At first, sales will increase at an increasing rate and then increase at a decreasing rate. The value of x where the rate of change of sales changes from increasing to decreasing is called the point of diminishing returns. This is also the point where the rate of change has a maximum value. Money spent after this point may increase sales, but at a lower rate. The next example illustrates this concept.

Currently, a discount appliance store is selling 200 large-screen television sets monthly. If the store invests \$x thousand in an advertising campaign, the ad company estimates that sales will increase to

N (x) = 3x3 – 0.25x4 + 200 0 <x< 9

When is rate of change of sales increasing and when is it decreasing? What is the point of diminishing returns and the maximum rate of change of sales?

Example(continued)

Solution:

The rate of change of sales with respect to advertising expenditures is

N ´(x) = 9x2 – x3 = x2(9 – x)

To determine when N ´(x) is increasing and decreasing, we find N´´(x), the derivative of N ´(x):

N ´´(x) = 18x – 3x2 = 3x(6 – x)

The information obtained by analyzing the signs of N ´(x) and N ´´(x) is summarized in the following table (sign charts are omitted).

Example(continued)

Example(continued)

Examining the table, we see that N ´(x) is increasing on (0, 6) and decreasing on (6, 9). The point of diminishing returns is x = 6, and the maximum rate of change is N ´(6) = 108. Note that N ´(x) has a local maximum and N (x) has an inflection point at x = 6.