Let a = 3, so f(x) = 3x. Evaluate f at 2, 1, 0, -1, -2 and use these points to sketch the graph of f.
| f(2) = 32 = 9 |
|
| f(1) = 31 = 3 | |
| f(0) = 30 = 1 | |
| f(-1) = 3-1 = 1/3 | |
| f(-2) = 3-2 = 1 / 32 = 1/9 |
f(x) = 3x |
Exponential Functions |
Contents: This page corresponds to § 4.1 (p. 316) of the text
Suggested problems from text:
p. 325 #9, 11, 17, 21, 24, 31, 39, 44, 53, 56, 62, 63, 64, 65, 67, 68
Introduction
Graphs
The Natural Base e
Compound Interest
So far, all of our exponents have been numbers. We are comfortable with constants like 32 and the function x3. In this section we start letting exponents be expressions with variables.
Let a be a positive number other than 1. The exponential function f with base a is defined by
f(x) = ax
Example 1.
Let a = 3, so f(x) = 3x. Evaluate f at 2, 1, 0, -1, -2 and use these points to sketch the graph of f.
f(2) = 32 = 9
f(1) = 31 = 3 f(0) = 30 = 1 f(-1) = 3-1 = 1/3 f(-2) = 3-2 = 1 / 32 = 1/9 f(x) = 3x
Example 2.
Let a = 1/2, so f(x) = (1/2)x. Evaluate f at 2, 1, 0, -1, -2 and use these points to sketch the graph of f.
First note that (1/2)x = 1 / 2x = 2-x, and the latter form is often more convenient.
f(2) = (1/2)2 = 1/4
f(1) = (1/2)1 = 1/2 f(0) = (1/2)0 = 1 f(-1) = 2-(-1) = 21 = 2 f(-2) = 2-(-2) = 22 = 4 f(x) = (1/2)x = 2-x
There are several things to notice about the graphs in the examples above.
The graphs of all exponential functions have these characteristics. They all contain the point (0, 1), because a0 = 1. The x-axis is always an asymptote. They are decreasing if 0 < a < 1, and increasing if 1 < a. The only thing that varies is the steepness.
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1 < a |
0 < a < 1 |
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f(x) = ax |
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Given a point in the upper half plane, i.e. above the x-axis, that is not on the y-axis, then there is a unique exponential function whose graph contains that point. ( Note that if the point lies on the horizontal line y = 1, then it lies on the graph of f(x) = 1x. This function is constant and hence not very interesting!) The applet below illustrates this fact. Initially the variable point is (1, 2), so the exponential function determined is f(x) = 2x, and this is indicated in the text box below the graph. As you drag the point, the function and graph are updated. Note that unlike earlier applets, here you cannot drag the point (0,1), because the graphs of all exponential functions contain that point!
Exponential function determined by a point.
Drag the point (1, 2) to update.
In general, equations involving exponential functions are easier to graph than polynomial equations because the exponential functions themselves are easy to graph.
Sketch the graphs of y = 2x, y = 2x + 3, and y = 2x + 3 - 2 in the same coordinate plane. Answer
There are infinitely many exponential functions, one for each positive number base. However, there is one that is more important than all the others, the exponential function base e. This function plays such a prominent role in mathematics and its applications, that it is often called simply the exponential function, as if there were not any others!
The number e is approximately 2.718281828459045.
It is not possible to give the exact value of e in decimal form because it is irrational. Its decimal expansion never terminates or repeats.
Any explanation of why e is so important will rely in some way on calculus. In these notes we will introduce you to an important property of the exponential function ex that concerns lines tangent to a graph.
Recall that the graph of every exponential function contains the point (0, 1). For each of these graphs there is a line through (0,1) that is tangent to the graph at the point. A line is tangent to a graph at a point if it is the "best straight line approximation" of the graph at that point. For graphs of exponential functions, the tangent line at a point intersects the graph only at that point.
The applet below shows the graph of an exponential function and the graph of the line tangent to the graph at the point (0,1). You can drag the point to produce another exponential graph and its tangent line. The formula for the exponential function and the slope of its tangent line are reported in the text boxes below the graph.
Exponential function and tangent line.
Drag the point to see other exponentials and their tangents.
One important property of the number e is that the line tangent to the graph of f(x) = ex at (0,1) has slope 1. With the applet above you cannot hit the graph of ex exactly, but you can come close, and when you do you see that the tangent slope is close to 1. In fact, when the applet first starts, the exponential function is quite close to ex.
All scientific calculators provide some means of evaluating the exponential function. For instance, on the TI-82 you would find e2 by pressing the 2nd key followed by the LN key, then 2 and enter. On the Java Calculator you evaluate e2 either by typing e^2 and enter, or exp(2) and enter. The answer should be close to 7.3890560989306495.
The following table explains the basic terms used in compound interest problems.
| P | Principal | The principal is the amount that you invest. |
| A | Compound Amount | The compound amount is what your investment has grown to. |
| r | Annual Interest Rate | This rate is often given as a percent and must be rewritten in decimal form. |
| n | Number of Compounding periods per year |
For example, when interest is compounded quarterly, there are four compounding periods in a year, so n = 4. |
| t | Number of years | The length of time (in years) for the investment. |
Compound Interest Formula
Notes:
Example 3.
If $1000 is invested at 8 % annual interest compounded monthly for 4 years, what does the investment grow to?
We want to find A. P = 1000, r = 0.08, t = 4, and n = 12 because we are compounding monthly and there are twelve months in a year.
If the interest rate stays the same and the number of compounding periods increases, then the yield from the investment is greater. The next exercise illustrates this fact.
(a) Show that if $1000 is invested at 8 % compounded quarterly for 4 years, the compound amount is $ 1372.79. Note that there are fewer compounding periods than in example 3, and the yield is less.
(b) Show that if $1000 is invested at 8 % compounded daily (n = 360) for 4 years, the compound amount is $ 1377.08. Note that there are more compounding periods than in example 3, and the yield is greater. (I know there are more than 360 days in a year, but 360 is the value commonly used for compounding daily.)
We could use more than 360 compounding periods in a year. we could consider compounding hourly or every minute, but the resulting formulas become very cumbersome, and the increase in yield in not that great.
Continuous Compounding is the limit of taking shorter and shorter compounding periods. Consider going to extremes with compounding periods. Compound every hour, every minute, every second, etc. As you continue this sequence, the formulas are getting worse and worse, but you are approaching continuous compounding whose formula is simpler than any of the compound interest formulas we have encountered!
Formula for Continuous Compounding
A = P ert
Notes:
Example 4.
$500 is to be invested at 7 % annual interest for 3 years.
If interest is compounded monthly, then the compound amount is
A = 500(1 + 0.07/12)^(12*3) = $ 616.46.
If interest is compounded continuously, then the compound amount is
A = 500 e0.07*3 = $ 616.84.
If $1000 is invested at 8 % annual interest compounded continuously for 4 years, show that the compound amount is $1377.13. Compare this with example 3 above.
