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
Determine whether the function
is continuous at x = 4.
A. yes
B. no
Determine whether the function
is continuous at x = 2.
A. yes
B. no
Describe the end behavior of
f(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
on
.
and increasing
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
Analyze Graphically
From the graph, we can estimate that f is decreasing on
, increasing on
, and decreasing on
.
Support Numerically
Create a table using x-values in each interval.
Analyze Increasing and Decreasing Behavior
Analyze Increasing and Decreasing Behavior
The table shows that as x increases to
decreases; as x increases from
as x increases from
, f(x)
, f(x) increases;
, f(x) decreases. This supports
the conjecture.
Answer: f(x) is decreasing on
and increasing on
and
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
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.
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.
Estimate and Identify Extrema of a Function
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.
f(100) < f(2) and f(–100) > f(–1), which supports our
conjecture that f has no absolute extrema.
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).
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.
Answer: There is a maximum of about 70 miles per
hour. The car will travel 362.5 miles when
traveling at the optimum speed.
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.
The average rate of change on the interval [–3, –1] is
12. The graph of the secant line supports this
conclusion.
Answer: 12
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.
The average rate of change on the interval [2, 5] is
–10. The graph of the secant line supports this
conclusion.
Answer: –10
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
Find Average Speed
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.
Substitute 1 for t1 and
2 for t2.
Evaluate d(2) and d(1).
Simplify.
Find Average Speed
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.
Answer: 8.1 feet per second
Find Average Speed
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.
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