[| Exponential Function |
Exponential Functions
The exponential function is used in many different fields of reality including banking (interest), biology research (culture growth), demographics (population growth) and physics (radioactive decay)
We use exponential functions to calculate depreciation on capital assets and interest on bank accounts and loans. Thus, it's obvious that there are 2 kinds of exponential functions -- growth curves and decay curves. In the first case, growth curves, something such as bacteria, a population or even your bank account balance gets bigger or grows. In the case of a decay curve, something such as a mass of radioactive material or the value of a truck gets smaller or decays.
Two things are always true of all exponential functions
There are 2 sets of exponential functions:
| f (x) = a(c) ( x – h ) + k | 0 < c < 1 (c is a fraction < 1) | |
| a > 0 always decreasing | domain: R | range: ] k, [ |
| a < 0 always increasing | domain: R | range: ] –º , k [ |
| f (x) = a(c) ( x – h ) + k | c > 1 | |
| a > 0 always increasing | domain: R | range: ] k,  [ |
| a < 0 always decreasing | domain: R | range: ] – , k [ |
When given f (x) = a(c) b ( x – h ) + k, raise c b for the new base to eliminate b.
example: f (x) = – 5 (3) 2 (x – 5 ) + 7 becomes f (x) = – 5 (9) (x – 5 ) + 7
Let's look at the two best known and some transformed exponential functions.
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| f(x) = 2 x (doubling), g(x) = (½) x (halving). | The transformed exponential functions below are f(x) = 3(0.25) x + 1 - 6, g(x) = – 2(3) x – 4 + 1, and h(x) = – (0.3) x – 1 + 5 |
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Once we decide where y = k is situated, the sign of a and the value of c will tell us if the curve increases or decrease over its domain.
Note: every exponential curve has a y-intercept found by setting x = 0.
There might be a zero depending on where the asymptote is located and
if the curve is increasing or decreasing.
If we consider the curve for g(x) above. Had k been – 1 instead of +1, the curve
would never cross the x-axis. The function would not have a zero.
(solving exponential and log equations and word problems on
this function are found in fct16: solving exponential and log equations)
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Practice
1) Using the coordinates of P and Q, find the rule for an exponential function
of the form f (x) = ac x which satisfies the situation.
a) P (0, 5) and Q (1, 15)
b) P (0, – 4) and Q (1, – 20) (click here to see solution)
2) find the rule for an exponential function of the form f (x) = ac x + k which
goes through points A (1, 7.4) and B(2, 5.72) if no y-value is less than or equal to 5.
(hint: set up 2 equations in 2 unknowns, then divide one by the other to find c)
(click here to see solution)
3) At a temperature of 10° C, an ice cube's volume decreases by 40% each hour.
If the volume of our ice cube is 20 cm3 before it starts to melt:
a) write the rule of correspondence that defines this situation.
b) what is the volume of the ice cube after 3 hours?
c) how many hours will it take for the volume to be less than 1 cm3 ?
(click here to see solution)
4) Ashley bought a car in 2000 for which she paid $19,500.
Three years later, the value of the car was $12,000.
If the depreciation curve is an exponential function,
how old will her car be when its value is less than $7000?
(click here to see solution)
5) The Mayors of two Québec towns were talking about the population of their respective cities. The Mayor of Sans Cerveau says that the present population of his town is 40,000 but experts have said that it will decrease by 3% per year over the next 30 years. The Mayor of St. Jean-de-Rubbershoe says that the present population is 10,000 but it will grow by 5% annually over the next 30 years. In how many years will the 2 towns have the same population? Round your answer to the nearest tenth of a year.
(click here to see solution)
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Solutions
1) f(x) = ac x
| a) P (0, 5) and Q (1, 15) substitute P so 5 = ac 0 so 5 = a so now f(x) = 5c x, we substitute Q 15 = 5(c) 1 so c = 3 f(x) = 5(3) x |
b) P (0, – 4) and Q (1, – 20) substitute P so – 4 = ac 0 so – 4 = a so now f(x) = – 4c x, we substitute Q – 20 = – 4(c) 1 so c = 5 f(x) = – 4(5) x . |
2) if no y-value is less than or equal to 5, k = 5.
f(x) = ac x + k so f(x) = ac x + 5
substitute the coordinates of A and B to get:
7.4 = ac + 5 so 2.4 = ac
5.72 = ac 2 + 5 so 0.72 = ac 2, and now we divide
and a = 8
f(x) = 8(0.3) x + 5
| a) V(t) = 20(0.6) t | b) V(3) = 20(0.6) 3 = 4.32 cm 3. | c) 1/20 = (0.6) t (take logs) log (0.05) = t log (0.6) t = log (0.05) / log (0.6) = 5.86 hr. or use ln instead of log: |
4) We have 2 points on the curve: (0, 19,500) and (3, 12,000)
So we know that V(t) = V0 (c) t so 12000 = 19500c 3 so 120/195 = c 3 so c = 0.85
Now set 7000 = 19500(0.85) t , use logs to solve for t.
You should get t = 6.3 years
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5) The rule of correspondence for Sans Cerveau is P(t) = 40,000(0.97) t .
The rule of correspondence for St. Jean-de-RubberShoe is P(t) = 10,000(1.05) t .
We want to solve for t that makes them equal.
40,000(0.97) t = 10,000(1.05) t so 4(0.97) t = (1.05) t (divide through by 10,000)
Now divide by (0.97) t to get:
The two towns will have the same population after 17.5 years.
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