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Math 229 Exponential Functions
A. Definition: An exponential function is a function of the form f (x) = ax^ for 0 < a < 1 or a > 1. Note: a is called the base of the exponential function.
The reason we exclude 0 and 1 as bases for exponential function is because 0x^ = 0 for and x, and 1x^ = 1 for any x, so these are just constant functions.
Example 1: Let f (x) = 3x. Then: (a) f (0) = 3^0 = 1 (b) f (2) = 3^2 = 9 (c) f (−3) = 3−^3 = 271
(d) f ( 23 ) = 3 (^23) = 3
Graphs of exponential functions:
f (x) = 3x x f (x) − 3 271 − (^2 ) − (^1 ) 0 1 1 3 2 9 3 27
y 10 9 8 7 6 5 4 3 2 1
−4 −3 −2 −1 1 2 3 4 x
f(x) = 3x
(0,1)
f (x) =
3
)x
x f (x) 3 271 (^2 ) (^1 ) 0 1 − 1 3 − 2 9 − 3 27
x 10 9 8 7 6 5 4 3 2 1
−4 −3 −2 −1 1 2 3 4 x
y
(0,1)
f(x) = (1/3)
Properties of Exponential Graphs:
- Domain: (−∞, ∞)
- Range: (0, ∞)
- y-intercept: (0, 1), x intercept: none.
- Continuous everywhere
- Increasing if b > 1. Decreasing if 0 < b < 1.
Solving Basic Exponential Equations:
Examples:
- 4^2 x−^3 = 4^5 −x Since f (x) = 4x^ is a one-to-one function, we can conclude that: 2 x − 3 = 5 − x, or 3x = 8. Hence x = 83.
- 2^4 x−^7 = 8^2 x−^5 Since 8 = 2^3 , we can rewrite 8^2 x−^5 as
) 2 x− 5 = 23(2x−5)^ = 2^6 x−^15. Then, as above, we know that 4x − 7 = 6x − 15, or 8 = 2x. Hence 4 = x.
The Compound Interest Formula: When a principal amount P in invested at interest rate r which is compounded n
times per year and remains invested for t year, the amount A that results is given by the formula A = P
r n
)nt
Examples:
- Suppose you put $1000 in an account that pays 6% interest compounded monthly. How much money will be in the account 3 years later?
P = 1000, r = 0.06, n = 12, and t = 3, so A = 1000
= 1000 (1.005)^36 ≈ $1, 196. 68
- Now Suppose you put $2000 in an account that pays 7% interest compounded daily. How much money will be in the account 5 years later?
P = 2000, r = 0.07, n = 365, and t = 5, so A = 2000
1 + 365.^07
≈ 2000 (1.000191781)^1825 ≈ $2838. 04
The Natural Exponential Function:
Definition: If we consider what happens to the base of our compound interest exponential term:
1 + (^1) n
as we compound more and more frequently n
1 + (^) n^1
)n
1 2. 10 2. 100 2. 1,000 2. 10,000 2. 100,000 2. 1,000,000 2.
Therefore, we define e = lim n→∞
n
)n
. We say e is the base of the natural exponential function.
Continuously Compounded Interest Using this new base, we can measure the accumulation of interest that is com- pounded “instantaneously” rather than only n times a year. We do so using the formula: A = P ert, where P, A, r, and t are exactly as above.
Example: Suppose you invest $1000 at 6% interest compounded continuously for 3 years. Then at the end of the 3 years, you will have: 1000e^0 .06(3)^ ≈ $1, 197. 22 Notice that this is about 54 cents more that we had investing the same amount at the same interest rate but only compounded monthly.
Example: Suppose the population of a bacterial colony if given by the function f (t) = 500e−.^87 t^ where t is in hours and f (t) is in thousands of cells. Then f (0) = 500e−.087(0)^ = 500e^0 = 500, so there are initially 500,000 cells in the colony. Similarly, f (5) = 500e−.087(5)^ = 500e−^0.^435 ≈ 323 .632, so after 5 hours, the population of the colony has been reduced to 323,632 cells. The Derivative of the Exponential Function: The basic rule for differentiating the exponential function is: d dx
ex^ = ex
We’ll not prove this rule, but it is in fact true that the exponential function is its own derivative!
The Chain Rule for Exponentials:
d dx
ef^ (x)^ = f ′(x) · ef^ (x)
Examples of Derivatives Involving Exponential Functions:
- If f (x) = e^3 x, then, using the Chain Rule for Exponentials: f ′(x) = e^3 x^ · (3) = 3e^3 x.
- If g(x) = ex 2 , then, using the Chain Rule for Exponentials: g′(x) = ex 2 · (2x) = 2xex 2 .
- If h(x) = x^2 e^5 x, then, by the product rule: h′(x) = 2xe^5 x^ + 5x^2 e^5 x.
- If k(x) =
e^2 x^ + 3x^2
e^2 x^ + 3x^2
2 e^2 x^ + 6x
5 e^2 x^ + 15x
e^2 x^ + 3x^2
- If ℓ(x) = ee
x^2 , then, applying the Chain Rule for Exponentials several times: ℓ′(x) = ee
x^2 · ex
2 · x^2
Applications of the Derivative Involving Exponentials:
- Find the slope of the tangent line to f (x) = 1 − e^2 x^ at the point where f crosses the x-axis. Then find the equation of the tangent line.
First notice that if f (x) = 0, then 0 = 1 − e^2 x, so e^2 x^ = 1, or, ln(e^2 x) = ln(1). Therefore, 2x = 0, so x = 0. Therefore, the point of tangency is (0, 0). Next, f ′(x) = − 2 e^2 x, so m = f ′(0) = − 2 e^0 = −2(1) = −2. Thus, the tangent line has equation y = − 2 x.
