Download 111 Ch2-4

2.4 Modeling with Linear Functions
To model a quantity that is changing at a constant rate with f(x) = ax + b, the
following formula may be used.
f(x) = (constant rate of change)x + (initial amount)
The constant rate of change corresponds to the slope of the graph of f, and the
initial amount corresponds to the y-intercept.
Example 2 (p.119): Finding a symbolic representation
A 100-gallon water tank, initially full of water, is being drained at a rate of 5
gallons per minute.
a) Write a formula for a linear function f that models the number of gallons of
water in the tank after x minutes.
Water in the tank is decreasing at a constant rate of 5 gal/min. The rate of
change is –5. The initial amount of water is 100 gal. The formula is
𝑓(𝑥) = −5𝑥 + 100
b) How much water is in the tank after 4 minutes?
Find 𝑓(4). 𝑓(4) = −5 ∙ 4 + 100 = −20 + 100 = 80
After 4 minutes there are 80 gallons of water in the tank.
c) Graph f. Identify the x- and y-intercepts and interpret each.
y-intercept is (0, 100): initially, the tank contains 100 gallons of water.
x-intercept is (20, 0): in 20 minutes the tank will be empty (0 gallons of water).
d) What is the domain of f ?
The domain of f must be restricted to {x | 0 ≤ x ≤ 20}.
You can’t have negative time.
The tank can’t hold a negative amount of water.
Piecewise-Defined Functions (p.121)
A function f that models data and is defined on pieces of its domain is called a
piecewise function. If the graph for each piece of the domain is linear, the function
is a piecewise-linear function.
The next example shows a piecewise-constant function or a step function.
Fujita Scale - Intensity of Tornadoes (p.121)
An F1 tornado has wind speeds between 40 and 72 miles per hour. Here is the
piecewise function for the F-scale.
1
2
𝑓(𝑥) = 3
4
{5
if
40 ≤ 𝑥 ≤ 72
if 72 < 𝑥 ≤ 112
if 112 < 𝑥 ≤ 157
if 157 < 𝑥 ≤ 206
if 206 < 𝑥 ≤ 260
Example 5 (p.123): Evaluating and graphing a piecewise function
𝑥 − 1 if − 4 ≤ 𝑥 < 2
Use 𝑓(𝑥) = {
to complete the following:
−2𝑥 if
2≤𝑥≤4
a) What is the domain of f ?
f is defined for –4 ≤ x < 2 or 2 ≤ x ≤ 4
The domain is D = {x |–4 ≤ x ≤ 4}, or [–4, 4]
b) Evaluate f(–3), f(2), f(4), and f(5).
f(–3) = –3 – 1 = –4
Use x – 1
f(2) = –2 • 2 = –4
Use –2x
f(4) = –2 • 4 = –8
Use –2x
f(5) is undefined (5 is not in the domain)
c) Sketch a graph of f.
From –4 up to 2, graph y = x – 1 with a solid dot at –4 and an open circle at 2
From 2 to 4, graph y = –2x with a solid dot at both 2 and 4
d) Is f a continuous function on its domain?
f is not continuous. There is a break at x = 2
Greatest Integer Function (p.123)
The greatest integer function 𝑓(𝑥) = ⟦𝑥⟧ is defined as follows.
⟦𝑥⟧ is the greatest integer less than or equal to x.
−2 if − 2 ≤ 𝑥 < −1
−1 if − 1 ≤ 𝑥 < −2
⟦𝑥⟧ =
0 if
0≤𝑥<1
1 if
1≤𝑥<2
{ 2 if
2≤𝑥<3
Be careful with negative values for x.
⟦−1.7⟧ = −2
NOT − 1
⟦−3.2⟧ = −4
NOT − 3
Direct Variation (p.124)
Let x and y denote two quantities. y is directly proportional to x, or y varies
directly with x, if there exists a nonzero number k such that
y = kx
k is called the constant of proportionality or the constant of variation.
Hooke’s Law (p.125)
Hooke’s Law states: The distance an elastic spring stretches beyond its natural
length is directly proportional to the amount of weight hung on the spring. This is
illustrated in the figure.
This law is valid whether the spring is
stretched or compressed.
The constant of proportionality is called
the spring constant.
If a weight or force F is applied and the spring
stretches a distance x beyond its natural length,
then the equation F = kx models this situation,
where k is the spring constant.
Example 6 (p.125): Working with Hooke’s Law
A 12-pound weight is hung on a spring, and stretches it 2 inches.
a) Find the spring constant.
F = kx where F = 12 pounds and x = 2 inches
12 = k(2) so k = 6
The spring constant is 6 and F = 6x.
b) Determine how far the spring will stretch when a 19-pound weight is hung on it.
For F = 19, we have 19 = 6𝑥
𝑥=
19
6
𝑥 ≈ 3.17
The spring will stretch about 3.17 inches.
Solving a Variation Problem (p.125)
When solving a variation problem, the following steps can be used.
STEP 1: Write the general equation for the type of variation problem that you are
solving.
STEP 2: Substitute given values in this equation so the constant of variation k is
the only unknown value in the equation. Solve for k.
STEP 3: Substitute the value of k in the general equation in Step 1.
STEP 4: Use this equation to find the requested quantity.
Example 7 (p.125): Solving a direct variation problem
Let T vary directly with x, and suppose that T = 33 when x = 5.
Find T when x = 31.
Solution:
STEP 1: Direct variation equation is T = kx
STEP 2: Substitute 33 for T and 5 for x , then solve for k
𝑇 = 𝑘𝑥
33 = 𝑘 ∙ 5
𝑘=
33
5
= 6.6
STEP 3: 𝑇 = 6.6𝑥
STEP 4: When x = 31, we have 𝑇 = 6.6 ∙ 31
𝑇 = 204.6