Week 8 Exponential Functions Many images below are excerpts from the multimedia textbook. You can find them there and in your textbook in sections 4.1 and 4.2. With the beginning of the new chapter we encounter exponential functions, a new class of functions. They are really different from the polynomial and rational functions we have studied up to this point. We need to realize that much of our mathematical intuition was shaped by our experience with polynomials and that it might fail us, dealing with exponential functions. Exponential Growth is synonymous with rapid growth in every day speech. It means that the growth can be modeled by an exponential function. See some examples below. Watch the video for section 4.2. Work through Example 1 in section 4.2 starting on page 298. You find the following graph for f(x) = 2 x. It is typical for any function of the form f(x) = a x with a>1.
We find the graph to have a y-intercept of 1, a function value of a when x = 1, and increasing rapidly for positive x-values. For negative x-values we see that the graph approaches the x-axis as horizontal asymptote. The function never crosses the x-axis and has no negative function values. The function is defined for all values of x and is continuous. The domain is (-, ) and the range is (0, ). The function is increasing for all values of x. f(x) = a x is in fact one of our new basic graphs. We need to learn to graph it by hand for all positive values of a. Work through Example 2 in section 4.2 starting on page 299. The next graph is typical for values of a satisfying 0<a<1. Here is the graph of f(x) = (1/2) x.
Discussion Question 1: Can you obtain the graph of f(x)=(1/2) x from the graph of f(x) = 2 x by a transformation? How? Explain. Discussion Question 2: Do the interactive discovery on page 300. Answer the questions: What relationship do you see between the base a and the shape of the resulting graph of a x? What do all the graphs have in common? How do they differ? Work through Example 3 in section 4.2 on page 301.
Discussion Question 3: Find an exponential function obtained by horizontal and vertical shifts from a function of the form f(x) = a x with horizontal asymptote y = 3 and a y-intercept of (0, 5). Explain how you went about solving this problem. Since exponential functions are so different from the polynomial functions we are used to, we have to investigate some of the notions we are using nearly without thinking. When we solve an equation of the form x 2 = 36 for x, we have to apply an operation which "undoes" the squaring. That is we have to apply the square-root function. The operations of squaring and of taking the square root are one example of functions, which are inverses of each other. The study of exponential functions requires us to solve exponential equations and therefore we need to be able to "undo" exponential functions. This forces us to study the relationships of functions, which "undo" each other in more detail. For this reason we are now taking a closer look at function composition and inverses. We will study the actual inverse of an exponential function in week 9 when we learn about logarithmic functions. Function Composition The composition of functions is another way to combine functions that means to create more complicated functions from simple ones. We have previously encountered addition and multiplication of functions. Watch the video for section 4.1. Work through Example 1 in section 4.1 on page 283.
Discussion Question 4: Explain how you find (fοg)(x) and (gοf)(x) in Example 1b in section 4.1 on page 283. Don't "do" the example, but describe the procedure. Work through Example 2 in section 4.1 on page 284. Discussion Question 5: Explain how you find the domain of (fοg)(x) and (gοf)(x) in Example 2 in section 4.1 on page 284. Don't "do" the example, but describe the procedure. Work through Example 3 in section 4.1 on page 285. Discussion Question 6: Explain how one can decompose the function 1 h(x)= 4 in two different ways as composition of two functions f(x) (x 2) and g(x), that is h(x) = (fοg)(x). Inverse Functions Some functions "undo" each other when they are composed with each other. If we think of a function as a set of pairs, consisting of inputs and corresponding outputs. Undoing that function would mean that the ordered pairs have to be in reversed order. Work through Example 4 in section 4.1 on page 286. We usually don't deal with a function as a set of ordered pairs unless we graph it. This means for the graphs of two functions, which are inverses of each other, that the roles of x and y are switched. This idea also gives us the clue how to find an inverse function algebraically:
We see that the graph of the inverse relation is the reflection of the original graph on the line y =x. This example also shows that the inverse relation is not always a function. The red curve does not satisfy the vertical line test, because there are some x-values, which are related to two y-values. For this reason we can only find inverse functions for those functions which satisfy the horizontal line test: The one-to-one functions are simply those functions, which satisfy the horizontal line test.
Discussion Question 7:Explain in your own words what f -1 means. Use a couple of examples to illustrate your explanations. Discussion Question 8: Work through Example 6 in section 4.1 on page 288. How do you prove that a function is one-to-one? Discussion Question 9: Work through Example 7 in section 4.1 on page 288. How do you prove that a function is not one-to-one?
We are now ready to give the rule for finding the inverse function for a one-to-one function: Work through Examples 9, 10 and 11 in section 4.1 starting on page 290. We see again in Example 11 on page 291 that the graph of the inverse function is the reflection of the graph of the original function on the line y = x. Now combining the ideas of composition and inverse we finally get to the main point: The functions f and f -1 "undo" each other if we compose them.
In some cases we start with a function which is not one-to-one like f(x)=x 2. We need to find an inverse any way, to be able to solve equations. We can achieve that, by restricting the domain of f(x) to the interval [0, ). The function is one-to-one on this interval and therefore has an inverse function: f -1 (x) = x. Inverses of Exponential Functions We now return to the exponential functions f(x) = a x for a>0. We found that all these functions are one-to-one, so they all have inverse functions.
We will study these inverse functions, the logarithmic functions, in more detail in our next unit.