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§10.2 Second Derivative and Graphs . (11.2)

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- 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.

Find:

- the intervals where f(x) is increasing.
- the intervals where f(x) is decreasing
- the local maximum values, if any.
- the local minimum values, if any

The term concave upward is used to describe a portion of a graph that opens upward. Concave downward is used to describe a portion of a graph that opens downward.

Concave down

Concave 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).

Geometrically, the graph is concave upward on (a, b) if it lies above its tangent line at (a, b),

and is concave downward at (a, b) if it lies below its tangent line at (a, b).

down

tangent

up

For y = f (x), the second derivativeof f, provided it exists, is

f “ (x) =

Other notations for f “ (x) are

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).

Example. Page 268, #50. Find the intervals where the graph of f is concave upward, the intervals where the graph of f is concave downward if

f (x) = x3 + 24x2 + 15x - 12.

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).

f (x) = x3 + 24x2 + 15x - 12.

f “ (x) = 6x + 48

f (x) is concave up when x > -8.

f (x) is concave down when x < -8 .

f (x)

f “ (x)

- 8

-25 ≤ x ≤ 20 and – 400 ≤ y ≤ 14,000

- 10 ≤ x ≤ 1 and – 2 ≤ y ≤ 6

In general, an inflection point is a point on the graph where the concavity changes from upward to downward or downward to upward. This means that f “ (x) 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

The theorem means that an inflection point can occur only at critical value of f ‘. But, 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 “ (c) changes sign at c. (i.e. f “ (c) = 0)

and c is in the domain of f.

f ‘ (x) increasing

f ‘ (x) decreasing

f ‘ (x) is constant

f is increasing

f ‘ (x) > 0

f is decreasing

f ‘ (x) < 0

f is constant

f ‘ (x) = 0

f is concave up

f “ (x) > 0

f concave down

f “ (x) < 0

Inflection point

f “ (x) = 0

(- , -1) (- 1, 3) (3, )

f ’ (x) + + + 0 - - - - - - 0 + + + + +

- 1

1

3

f (x) increasing decreasing increasing

f (x) Maximum minimum

f “ (x) - - - - - - - 0 + + + + + + + +

f (x) concave down - inflect - concave up

Page 268, #50.Find the inflection points of f (x) = x3 + 24x2 + 15x - 12.

f ‘ (x) =

f ‘ (x) = 3x2 + 48x + 15

f “ (x) =

6x + 48 with a critical value at

x = - 8.

From the previous example we saw that f “ (x) was negative to the left of – 8 and positive to the right of – 8, so at x = - 8, f “ (x) is zero and an inflection point.

Sign charts for f ‘ (x) and f “ (x) may be useful.

continued

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

f ‘ (x) = 3x2 + 48x + 15

f “ (x) = 6x + 48 with a critical value (inflection point) at x = -8.

Sign charts for f “ (x) may be useful.

f “ (x) - - - 0 + + + + + + + +

continued

-25 ≤ x ≤ 20 and - 400 ≤ y ≤ 14,000

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

Graphing Calculators. 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.

Graphing the second derivative and using a root approximation routine shows the inflection point at – 8 quite easily.

continued

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

Graphing the second derivative and using a root approximation routine shows the inflection point at – 8 quite easily.

f “ (x) = 6x + 48

- 8

- 10 ≤ x ≤ 1 and – 2 ≤ y ≤ 6

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

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

The derivative is needed.

N ‘ (x) = -6x2 + 180x –750.

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

15

NOTE: This is the derivative of N (x)!

continued

- 5 ≤ x ≤ 50 and – 1000 ≤ y ≤ 1000

15

NOTE: This is the derivative N ‘ (x).

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

c. What is the maximum rate of change of sales?

We want the maximum for the derivative.

N ‘ (x) =

N ‘ (x) = -6x2 + 180x –750.

Maximum at x = 15.

N ‘ (15) =

600.

0 ≤ x ≤ 70 and –0.03 ≤ y ≤ 0.015

15

15

NOTE: This is N (x).

NOTE: This is N “ (x).

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

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

N ‘ (x) =

N ‘ (x) = -6x2 + 180x –750.

N “ (x) =

N “ (x) = -12x + 180

Critical value at x = 15.

continued

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 + 2000 <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?

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).

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.

- We can use the second derivative to determine when a function is concave up or concave down.
- When the second derivative is zero, we may get an inflection point in f (x) (a change in concavity).
- The second derivative test may be used to determine if a point is a local maximum or minimum.
- The value of x where the rate of change changes from increasing to decreasing is called the point of diminishing returns.

§11.3; 1, 3, 4, 5, 6, 7, 9, 11, 13, 15, 19, 21, 25, 29, 33, 37, 41, 43, 47, 51, 55, 59, 63, 67, 71,73.