1-unit growth and decay factors
Module 4 : Investigation 3
MAT 170 | Precalculus
September 26, 2016
exponential functions
Recall that an exponential function is a function of the form :
f(x) = abx
where a and b are real numbers, with b > 0.
2
exponential functions
Recall that an exponential function is a function of the form :
f(x) = abx
where a and b are real numbers, with b > 0.
We call a the initial value, and b the base.
3
questions 1 & 4
The function f(x) = 2x is an example of an exponential function.
2(a) If we allow x to vary, and it increases by 1, what happens to the
output of the function ?
2(b) When x varies by 1, the new output is
% of the old
output value. What is the percent change in the output value ?
2(c) If we allow x to vary, and it increases by 3, what happens to the
output of the function ?
2(d) If we allow x to vary, and it increases by 8, what happens to the
output of the function ?
(4) Give 3 more examples of exponential functions whose values
double whenever x increases by 1.
4
1-unit growth and decay factor
Let f(x) = abx where a and b are real numbers, with b > 0.
5
1-unit growth and decay factor
Let f(x) = abx where a and b are real numbers, with b > 0.
How does the value of f change over an interval of length ℓ ?
6
1-unit growth and decay factor
Let f(x) = abx where a and b are real numbers, with b > 0.
How does the value of f change over an interval of length ℓ ?
ending value (= f(x1 + ℓ))
starting value (= f(x1 ))
7
1-unit growth and decay factor
Let f(x) = abx where a and b are real numbers, with b > 0.
How does the value of f change over an interval of length ℓ ?
ending value (= f(x1 + ℓ))
abx1 +ℓ
=
starting value (= f(x1 ))
abx1
8
1-unit growth and decay factor
Let f(x) = abx where a and b are real numbers, with b > 0.
How does the value of f change over an interval of length ℓ ?
ending value (= f(x1 + ℓ))
abx1 +ℓ
=
= bx1 +ℓ−x1
starting value (= f(x1 ))
abx1
9
1-unit growth and decay factor
Let f(x) = abx where a and b are real numbers, with b > 0.
How does the value of f change over an interval of length ℓ ?
ending value (= f(x1 + ℓ))
abx1 +ℓ
=
= bx1 +ℓ−x1 = bℓ .
starting value (= f(x1 ))
abx1
10
1-unit growth and decay factor
Let f(x) = abx where a and b are real numbers, with b > 0.
How does the value of f change over an interval of length ℓ ?
ending value (= f(x1 + ℓ))
abx1 +ℓ
=
= bx1 +ℓ−x1 = bℓ .
starting value (= f(x1 ))
abx1
ℓ-unit growth/decay factor : bℓ
In other words,
f(x1 + ℓ) = bℓ · f(x1 ).
11
1-unit growth and decay factor
What is the percent change of f over an interval of length ℓ ?
12
1-unit growth and decay factor
What is the percent change of f over an interval of length ℓ ?
change in value (= f(x1 + ℓ) − f(x1 ))
1% of starting value (= f(x1 )/100)
13
1-unit growth and decay factor
What is the percent change of f over an interval of length ℓ ?
change in value (= f(x1 + ℓ) − f(x1 ))
abx1 +ℓ − abx1
=
1% of starting value (= f(x1 )/100)
abx1 /100
14
1-unit growth and decay factor
What is the percent change of f over an interval of length ℓ ?
change in value (= f(x1 + ℓ) − f(x1 ))
abx1 +ℓ − abx1
=
1% of starting value (= f(x1 )/100)
abx1 /100
( x1 +ℓ
)
abx1
ab
= 100
− x1
abx1
ab
15
1-unit growth and decay factor
What is the percent change of f over an interval of length ℓ ?
change in value (= f(x1 + ℓ) − f(x1 ))
abx1 +ℓ − abx1
=
1% of starting value (= f(x1 )/100)
abx1 /100
( x1 +ℓ
)
abx1
ab
= 100
− x1
abx1
ab
= 100(bx1 +ℓ−x1 − 1)
16
1-unit growth and decay factor
What is the percent change of f over an interval of length ℓ ?
change in value (= f(x1 + ℓ) − f(x1 ))
abx1 +ℓ − abx1
=
1% of starting value (= f(x1 )/100)
abx1 /100
( x1 +ℓ
)
abx1
ab
= 100
− x1
abx1
ab
= 100(bx1 +ℓ−x1 − 1)
= 100(bℓ − 1).
17
1-unit growth and decay factor
What is the percent change of f over an interval of length ℓ ?
change in value (= f(x1 + ℓ) − f(x1 ))
abx1 +ℓ − abx1
=
1% of starting value (= f(x1 )/100)
abx1 /100
( x1 +ℓ
)
abx1
ab
= 100
− x1
abx1
ab
= 100(bx1 +ℓ−x1 − 1)
= 100(bℓ − 1).
ℓ-unit percent change : 100(bℓ − 1) %.
18
1-unit growth and decay factor
What is the percent change of f over an interval of length ℓ ?
change in value (= f(x1 + ℓ) − f(x1 ))
abx1 +ℓ − abx1
=
1% of starting value (= f(x1 )/100)
abx1 /100
( x1 +ℓ
)
abx1
ab
= 100
− x1
abx1
ab
= 100(bx1 +ℓ−x1 − 1)
= 100(bℓ − 1).
ℓ-unit percent change : 100(bℓ − 1) %.
In other words, the percentage change from f(x1 ) to f(x1 + ℓ) is
100(bℓ − 1) %.
19
question 9
The following table shows values for an exponential function f.
x
f(x)
0
16
1
4
2
1
3
0.25
(a) What is the 1-unit decay factor for f ? Explain the meaning of this
value.
(b) What is the 1-unit percent change for f ? Explain the meaning of
this percentage.
(c) What is the initial value of f (i.e. the value of f when x = 0) ?
(d) Use the information from (a)-(c) to determine a function formula
for f.
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question 12
The following is the graph of an exponential function f.
(a) What is the 1-unit
decay factor for f ?
(b) What is the 1-unit
percent change for f ?
(c) What is the initial
value of f ?
(d) Use the information
from (a)-(c) to
determine a function
formula for f.
21
question 14
Let g(p) = 1.578(0.68)p .
(a) What does the number 1.578 represent for this function ?
(b) What does the number 0.68 represent for this function ?
(c) Whenever p increases by 1, the new output value will be
% of the old output value.
(d) What is the 1-unit percent change for g, and what does this
number tell us ?
22
question 15
An investment of $3300 increases by 4.7% each month.
(a) What is the 1-month percent change in the investment value ?
(b) When the time elapsed since the investment was made increases
by 1 month, the new value of the investment is
% of the old
value.
(c) What is the 1-month growth or decay factor and what does this
value tell us about the situation ?
(d) Write a function formula to model the value of the investment (in
dollars) in terms of the time elapsed since the investment was made
(in months).
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