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Key Concept 1

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Key Concept 1. Analyze Increasing and Decreasing Behavior. A. Use the graph of the function f ( x ) = x 2 – 4 to estimate intervals to the nearest 0.5 unit on which the function is increasing, decreasing, or constant. Support the answer numerically. Example 1. Analyze Graphically - PowerPoint PPT Presentation

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Page 1: Key Concept 1
Page 2: Key Concept 1

Analyze Increasing and Decreasing Behavior

A. Use the graph of the function f (x) = x 2 – 4 to

estimate intervals to the nearest 0.5 unit on which the function is increasing, decreasing, or constant. Support the answer numerically.

Page 3: Key Concept 1

Analyze Increasing and Decreasing Behavior

Analyze Graphically

From the graph, we can estimate that f is decreasing on and increasing on .

Support Numerically

Create a table using x-values in each interval.

The table shows that as x increases from negative values to 0, f (x) decreases; as x increases from 0 to positive values, f (x) increases. This supports the conjecture.

Page 4: Key Concept 1

Analyze Increasing and Decreasing Behavior

Answer: f (x) is decreasing on and increasing on .

Page 5: Key Concept 1

Analyze Increasing and Decreasing Behavior

B. Use the graph of the function f (x) = –x 3 + x to

estimate intervals to the nearest 0.5 unit on which the function is increasing, decreasing, or constant. Support the answer numerically.

Page 6: Key Concept 1

Analyze Increasing and Decreasing Behavior

Support Numerically

Create a table using x-values in each interval.

Analyze Graphically

From the graph, we can estimate that f is decreasing on

, increasing on , and decreasing on

.

Page 7: Key Concept 1

Analyze Increasing and Decreasing Behavior

0.5 1 2 2.5 3

–6 –13.125 –24

Page 8: Key Concept 1

Analyze Increasing and Decreasing Behavior

The table shows that as x increases to , f (x)

decreases; as x increases from , f (x) increases;

as x increases from , f (x) decreases. This supports

the conjecture.

Answer: f (x) is decreasing on and

and increasing on

Page 9: Key Concept 1

Use the graph of the function f (x) = 2x 2 + 3x – 1 to

estimate intervals to the nearest 0.5 unit on which the function is increasing, decreasing, or constant. Support the answer numerically.

A. f (x) is increasing on (–∞, –1) and (–1, ∞).

B. f (x) is increasing on (–∞, –1) and decreasing on (–1, ∞).

C. f (x) is decreasing on (–∞, –1) and increasing on (–1, ∞).

D. f (x) is decreasing on (–∞, –1) and decreasing on (–1, ∞).

Page 10: Key Concept 1
Page 11: Key Concept 1

Estimate and Identify Extrema of a Function

Estimate and classify the extrema to the nearest 0.5 unit for the graph of f (x). Support the answers numerically.

Page 12: Key Concept 1

Estimate and Identify Extrema of a Function

Analyze Graphically

It appears that f (x) has a relative minimum at

x = –1 and a relative maximum at x = 2. It also appears

that so we conjecture

that this function has no absolute extrema.

Page 13: Key Concept 1

Estimate and Identify Extrema of a Function

Because f (–1.5) > f (–1) and f (–0.5) > f (–1), there is a relative minimum in the interval (–1.5, –0.5) near –1. The approximate value of this relative minimum is f (–1) or –7.0.

Support Numerically

Choose x-values in half unit intervals on either side of the estimated x-value for each extremum, as well as one very large and one very small value for x.

Page 14: Key Concept 1

Estimate and Identify Extrema of a Function

f (100) < f (2) and f (–100) > f (–1), which supports our conjecture that f has no absolute extrema.

Likewise, because f (1.5) < f (2) and f (2.5) < f (2), there is a relative maximum in the interval (1.5, 2.5) near 2. The approximate value of this relative maximum is f (2) or 14.

Answer: To the nearest 0.5 unit, there is a relative minimum at x = –1 and a relative maximum at x = 2. There are no absolute extrema.

Page 15: Key Concept 1

Estimate and classify the extrema to the nearest 0.5 unit for the graph of f (x). Support the answers numerically.

A. There is a relative minimum of 2 at x = –1 and a relative maximum of 1 at x = 0. There are no absolute extrema.

B. There is a relative maximum of 2 at x = –1 and a relative minimum of 1 at x = 0. There are no absolute extrema.

C. There is a relative maximum of 2 at x = –1 and no relative minimum. There are no absolute extrema.

D. There is no relative maximum and there is a relative minimum of 1 at x = 0. There are no absolute extrema.

Page 16: Key Concept 1

Use a Graphing Calculator to Approximate Extrema

GRAPHING CALCULATOR Approximate to the nearest hundredth the relative or absolute extrema of f (x) = x

4 – 5x 2 – 2x + 4. State the x-value(s)

where they occur.

f (x) = x 4 – 5x

2 – 2x + 4

Graph the function and adjust the window as needed so that all of the graph’s behavior is visible.

Page 17: Key Concept 1

Use a Graphing Calculator to Approximate Extrema

From the graph of f, it appears that the function has one relative minimum in the interval (–2, –1), an absolute minimum in the interval (1, 2), and one relative maximum in the interval (–1, 0) of the domain. The end behavior of the graph suggests that this function has no absolute extrema.

Page 18: Key Concept 1

Use a Graphing Calculator to Approximate Extrema

Using the minimum and maximum selection from the CALC menu of your graphing calculator, you can estimate that f(x) has a relative minimum of 0.80 at x = –1.47, an absolute minimum of –5.51 at x = 1.67, and a relative maximum of 4.20 at x = –0.20.

Page 19: Key Concept 1

Use a Graphing Calculator to Approximate Extrema

Answer: relative minimum: (–1.47, 0.80); relative maximum: (–0.20, 4.20);absolute minimum: (1.67, –5.51)

Page 20: Key Concept 1

GRAPHING CALCULATOR Approximate to the nearest hundredth the relative or absolute extrema of f (x) = x

3 + 2x 2 – x – 1. State the x-value(s) where they

occur.

A. relative minimum: (0.22, –1.11);relative maximum: (–1.55, 1.63)

B. relative minimum: (–1.55, 1.63); relative maximum: (0.22, –1.11)

C. relative minimum: (0.22, –1.11);relative maximum: none

D. relative minimum: (0.22, 0); relative minimum: (–0.55, 0)relative maximum: (–1.55, 1.63)

Page 21: Key Concept 1

Use Extrema for Optimization

FUEL ECONOMY Advertisements for a new car claim that a tank of gas will take a driver and three passengers about 360 miles. After researching on the Internet, you find the function for miles per tank of gas for the car is f (x) = 0.025x

2 + 3.5x + 240, where x is the speed in miles per hour . What speed optimizes the distance the car can travel on a tank of gas? How far will the car travel at that optimum speed?

Page 22: Key Concept 1

Use Extrema for Optimization

We want to maximize the distance a car can travel on a tank of gas. Graph the function f (x) = –0.025x

2 + 3.5x + 240 using a graphing calculator. Then use the maximum selection from the CALC menu to approximate the x-value that will produce the greatest value for f (x).

Page 23: Key Concept 1

Answer: The optimal speed is about 70 miles per hour. The car will travel 362.5 miles when traveling at the optimum speed.

Use Extrema for Optimization

The graph has a maximum of 362.5 for x ≈ 70. So the speed that optimizes the distance the car can travel on a tank of gas is 70 miles per hour. The distance the car travels at that speed is 362.5 miles.

Page 24: Key Concept 1

VOLUME A square with side length x is cut from each corner of a rectangle with dimensions 8 inches by 12 inches. Then the figure is folded to form an open box, as shown in the diagram. Determine the length and width of the box that will allow the maximum volume.A. 6.43 in. by 10.43 in.

B. 4.86 in. by 8.86 in.

C. 3 in. by 7 in.

D. 1.57 in. by 67.6 in.

Page 25: Key Concept 1
Page 26: Key Concept 1

Find Average Rates of Change

A. Find the average rate of change of f (x) = –2x

2 + 4x + 6 on the interval [–3, –1].

Use the Slope Formula to find the average rate of change of f on the interval [–3, –1].

Substitute –3 for x1 and –1 for x2.

Evaluate f(–1) and f(–3).

Page 27: Key Concept 1

Find Average Rates of Change

Simplify.

Answer: 12

The average rate of change on the interval [–3, –1] is 12. The graph of the secant line supports this conclusion.

Page 28: Key Concept 1

Find Average Rates of Change

B. Find the average rate of change of f (x) = –2x

2 + 4x + 6 on the interval [2, 5].

Use the Slope Formula to find the average rate of change of f on the interval [2, 5].

Substitute 2 for x1 and 5 for x2.

Evaluate f(5) and f(2).

Page 29: Key Concept 1

Find Average Rates of Change

Simplify.

Answer: –10

The average rate of change on the interval [2, 5] is –10. The graph of the secant line supports this conclusion.

Page 30: Key Concept 1

Find the average rate of change of f (x) = –3x

3+ 2x + 3 on the interval [–2, –1].

A. 27

B. 11

C.

D. –19