bcb12e ppt 5 2

Chapter 5

Graphing and

Optimization

Section 2

Objectives for Section 5.2

Second Derivatives and Graphs

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

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The student will be able to find

inflection points.

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

Concavity

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

Definition of Concavity

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.

Concavity Tests

Theorem. 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:

Theorem. 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

′′

y

=

f

′′

(

x

) =

d

2

f

Example 1

Find the intervals where the graph of

Example 1

Find the intervals where the graph of

f (x) = x3 _{+ 24x}2 _{+ 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, ).

Example 1

(continued)

–10 < x < 1 –2 < y < 6 –25 < x < 20,

– 400 < y <14,000

f (x) _{f (x)}

Inflection Points

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.

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

Summary

f (x) > 0 f is increasing f (x) < 0 f is decreasing f (x) = 0 f is constant f (x) > 0 f (x) increasing f is concave up f (x) < 0 f (x) decreasing f concave down f (x) = 0 f (x) is constant f is linear Assume that f satisfies one of the conditions in the table below, for all x in some interval (a,b). Then the other

Example 2

Example 2

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.

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.

Example 2

(continued)

-8

–10 < x < 1 – 2 < y < 6 f ’’(x) = 6x + 48

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

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

Second Derivative Test - Concavity

f (c) f (c) graph of f is f (c) is

0

+

concave up local minimum

0

–

concave down local maximum

Curve Sketching

Graphing Strategy

 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).

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

Graphing Strategy

Example

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

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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 (3 + 45)/2.

■

Step 2. Analyze f (x).

f (x) = x2 – 2x – 3 = (x+1)(x–3), so f has critical values at –1 and 3.

Sign chart for

f



and

f



f (x) - - - 0 + + + + + + + +

- 1 1 3

(– , –1) (–1, 3) (3, )

f ’ (x) + + + 0 - - - 0 + + + + +

f (x) increasing decreasing increasing

f (x) maximum minimum

Analyzing Graphs - Applications

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

Analyzing Graphs - Applications

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

–5 < x < 50 and –1000 < y < 1000

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

Decreasing?

Note: This is the graph of the N (x) = -6x2 + 180x –750.

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

Application

(continued)

Application

(continued)

(b) Find the inflection points for the graph of N.

N ’(x) = –6x2 + 180x –750.

N ’’(x) = –12x + 180

Inflection point at x = 15.

0 < x < 70 and –0.03 < y < 0.015

Note: This is N (x).

15

Application

(continued)

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).

Point of Diminishing Returns

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

Maximum Rate of Change Example

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 _{– 0.25x}4 _{+ 200 0 < x < 9}

When is rate of change of sales increasing and when is it

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):

Example

(continued)

x N (x) N (x) N (x) N (x)

0 < x < 6

+

+

Increasing Increasing, concave up

x = 6

₀

+

Local Max Inflection Point

Example

(continued)

Examining the table, we see that N (x) is increasing on (0, 6) and decreasing on (6, 9). The point of

Summary

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We can use the second derivative to determine when a function is concave up or concave down.

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When the second derivative is zero, we may get an inflection point in f (x) (a change in concavity).

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The second derivative test may be used to determine if a point is a local maximum or minimum.

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The value of x where the rate of change changes from increasing to decreasing is called the point of