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These lecture notes from math 1300 cover the topic of exponential functions, including their definition, the relationship between initial value, growth factor, and percent change, the family of exponential functions, continuous growth and decay, and half-life and doubling time. Examples are provided to illustrate the concepts.
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MATH 1300 Lecture Notes Tuesday, August 27, 2013
(a) Exponential Functions: An exponential function has the form y = abx, or using function notation, f (x) = abx. Note that the variable, x, is in the exponent.
t t 0 (note that f (t 0 ) = 12 a, which is the definition of half-life). For example, if the half-life of a radioactive substance was 2 years, then the amount of this substance left after time t would be f (t) = a(^12 ) t 2 = a(.71)t. So the decay factor is .71 (meaning the substance is decreasing by 29% each year).
For example, if the doubling time of a population was 2 years, then population after time t would be f (t) = a(2) t 2 = a(1.41)t. So the growth factor is 1. 41 (meaning the population is increasing by 41% each year). (e) Concavity: Intuitively, concavity describes the way a graph bends on a given interval. If the graph bends upward from left to right on an interval (∪), we say it is concave up on that interval. If the graph bends downward on an interval (∩), we say it is concave down on that interval. For example, exponential functions are concave up everywhere: exponential growth functions resemble the right half of ∪, while exponential decay functions resemble the left half of ∪.
(a) Suppose the value of a brand new car is 25, 000 dollars. Write a function for the value of the car after t years if
Since the dependent variable (value of the car) is changing by a constant amount every year, this is describing a linear relationship. Since the ini- tial value is 25, 000 dollars and the value decreases 300 dollars per year, the function describing this relationship is
A(t) = − 300 t + 25, 000.
Since the dependent variable (value of the car) is changing by a constant percent every year, this is describing an exponential relationship. Since the initial value is 25, 000 dollars and the value decreases 12% every year (so the decay factor is 1 − .12 = .88), the function describing this relationship is
A(t) = 25, 000(.88)t.
(b) Suppose that g(x) is an exponential function with g(2) = 4 and g(5) = 20. Write the equation for g(x).
Solution. Since g(x) is exponential, we can write 4 = ab^2 and 20 = ab^5. Then taking the ratio, we get 20 4
ab^5 ab^2
so the a’s cancel and we get 5 = b^3 , so b = 3
4 = a( 3
so a = 4 5 23
. Since 3
5 ≈ 1 .71 and 4 5 23
≈ 1 .37, we have
g(x) = 1.37(1.71)x.