Embed Size (px)
Citation preview
Five-Minute Check
Then/Now
New Vocabulary
Key Concept: Increasing, Decreasing, and Constant Functions
Example 1: Analyze Increasing and Decreasing Behavior
Key Concept: Relative and Absolute Extrema
Example 2: Estimate and Identify Extrema of a Function
Example 3: Real-World Example: Use a Graphing Calculator to Approximate Extrema
Example 4: Use Extrema for Optimization
Key Concept: Average Rate of Change
Example 5: Find Average Rates of Change
Example 6: Real-World Example: Find Average Speed
Determine whether the function y = x 2 + x – 5 is
continuous at x = 7.
A. yes
B. no
A. yes
B. no
Determine whether the function is continuous at x = 4.
A. yes
B. no
Determine whether the function is continuous at x = 2.
Describe the end behavior off (x) = –6x 4 + 3x 3 – 17x
2 – 5x + 12.
A.
B.
C.
D.
Determine between which consecutive integers the real zeros of f (x) = x
3 + x 2 – 2x + 5 are located
on the interval [–4, 4].
A. [–2, –1]
B. [–3, –2]
C. [0, 1]
D. [–4, –3]
You found function values. (Lesson 1-1)
• Determine intervals on which functions are increasing, constant, or decreasing, and determine maxima and minima of functions.
• Determine the average rate of change of a function.
• increasing
• decreasing
• constant
• maximum
• minimum
• extrema
• average rate of change
• secant line
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.
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.
Analyze Increasing and Decreasing Behavior
Answer: f (x) is decreasing on and increasing on .
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.
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
.
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
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, ∞).
Estimate and Identify Extrema of a Function
Estimate and classify the extrema for the graph of f (x). Support the answers numerically.
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.
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 maximum 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.
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 minimum 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.
Estimate and classify the extrema 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.
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.
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), one relative 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.
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, a relative minimum of –5.51 at x = 1.67, and a relative maximum of 4.20 at x = –0.20.
Use a Graphing Calculator to Approximate Extrema
Answer: relative minima: (–1.47, 0.80); relative maximum: (–0.20, 4.20);absolute minima: (1.67, –5.51)
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)
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?
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).
Answer: There is a maximum of 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 ≈ 7.0. 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.
VOLUME A square with side length x is cut from each corner of a rectangle with dimensions 8 inches by 12 inches, then 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.
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).
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.
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).
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.
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
A. GRAVITY The formula for the distance traveled by falling objects on the Moon is d (t) = 2.7t
2, where d (t) is the distance in feet and t is the time in seconds. Find and interpret the average speed of the object for the time interval of 1 to 2 seconds.
Find Average Speed
Substitute 1 for t1 and 2 for t2.
Evaluate d(2) and d(1).
Simplify.
The average rate of change on the interval is 8.1 feet per second. Therefore, the average speed of the object in this interval is 8.1 feet per second.
Find Average Speed
Answer: 8.1 feet per second
B. GRAVITY The formula for the distance traveled by falling objects on the Moon is d (t) = 2.7t
2, where d (t) is the distance in feet and t is the time in seconds. Find and interpret the average speed of the object for the time interval of 2 to 3 seconds.
Find Average Speed
Substitute 2 for t1 and 3 for t2.
Evaluate d(3) and d(2).
Simplify.
Find Average Speed
The average rate of change on the interval is 13.5 feet per second. Therefore, the average speed of the object in this interval is 13.5 feet per second.
Answer: 13.5 feet per second
PHYSICS Suppose the height of an object thrown upward from the roof of a 50 foot building is given by h (t) = –16t
2 + 50, where t is the time in seconds after the object is thrown. Find and interpret the average speed of the object for the time interval 0.5 to 1 second.
A. 8 feet per second
B. 12 feet per second
C. 24 feet per second
D. 132 feet per second
