VIEWS: 0 PAGES: 10 POSTED ON: 10/1/2012 Public Domain
A Function & It’s Inverse Before we can discuss an inverse we must discuss one-to-one functions. Recall in Ch. 2.2 we discussed functions and we found that if there were no repeated x-values with different y-values (passes the vertical line test) that a relation was a function. The continuation of this is one-to-one relations and ultimately one-to-one functions. A One-to-One Relation has no repeated y-values with different x-values. In other words a one-to one relation passes a horizontal line test. We can use all our tricks from chapter 2 to tell if a relation is one-to-one. Example: Which relations are one-to-one? a) {(2,5), (2,6), (2,7)} Note: This isn’t a function. b) y = x, {x| x 0, x} Note: This is a function. c) 5 x -15 15 -5 Note: This isn’t a function. Now, the reason that we are concerned with one-to-one functions is our next concept. In order for a function to have an inverse it must be one-to-one and a function. The inverse for a set of ordered pairs can be found by exchanging the x and y coordinates (domain, range). Example: {(1,2),(2,3),(3,4),(4,5)} is a one-to-one function Y. Butterworth Ch. 9 Summary 1 For a function, the inverse function is denoted f –1(x) for the one-to-one function f(x). A function and its inverse are symmetric about the line y = x (you may recall that this is the line through the origin that divides the 1st and 3rd quadrants in half). The following is the procedure for finding the inverse of a function. Finding the Inverse of a One-to-One Function 1) Let x = y and f(x) = x 2) Solve for y 3) Replace y with f -1 (x) Example: Find the inverse function for f(x) = x 3, for x 3 then graph both functions on the same axis. Draw in the line y = x for reference. y x Note: The points on the function and its inverse are equidistant from the line y = x. Also note the x & y values are reversed! Y. Butterworth Ch. 9 Summary 2 An exponential function is any positive not equal to one raised to some valued exponent. F(x) = ax where a > 0, but a 1 The domain of the function, the values for which x can equal, are all (-,). Why? Because an exponent can be any real number. And the exponent is the independent variable. The range however, is all positive greater than zero (0, ). Why? Because any number raised to some power will always be a positive number. For all exponential functions, the graph will pass through the points: (-1,1/a) Why? That’s the reciprocal! (0,1) Why? Anything to zero power is 1 (1, a) Why? Anything to 1st power is itself. The shape of the curve will always be: (-1,1/a) (0,1) (-1,1/a) (1,a) (0,1) (1,a) Where a > 1 Where 0< a < 1 Increasing Function Decreasing Function Note: This can be when a>1, but x is neg. Just some notes: 1) An exponential graph will always approach the x-axis but will never touch the axis. 2) The graph will grow larger and larger if the number (a) is greater than 1. This is like population. Think of 2 people, they have children, then those have children and so on, this is an exponential function, and you should be able to see that it grows to an infinite number. 3) If the a is between one and zero (fractional) the shape is the same but starts at infinity and decreases to approach the x-axis. Note: If we have a –x, we are looking at the reciprocal of a to the x, so it will look like the graph of the function when 0 < a < 1, when a is greater than 1, and when a is between 0 and 1 the graph will look like the graph when a > 1. Exponential functions are all over in the world. Some of the most prominent areas are economics, biology, archeology, and some cross these boundaries into our everyday lives. One of the most prevalent cases is that of compounded interest. We would like to believe (based upon all interest problems that we have computed) that interest is simple interest, but it is generally compound, which increases at an exponential rate. Even the value of our vehicles once we drive them off the lot can be determined by an exponential function! Y. Butterworth Ch. 9 Summary 3 While we are doing graphing we will also graph the exponential function’s inverse, the logarithmic function. This means that the domain of the exponential becomes the range of the logarithmic and range becomes domain. It also means that the graph is symmetric to the exponential across the line y = x. Graphing a Log Function y = log a x a>0 &a1 converts to exponential x = ay 1) The domain is (0, ). 2) The range is (-,). 3) The points that are being plotted are (1/a, -1), (1,0) and (a,1) Note: The plot is the inverse of the exponential, the domains and ranges are flip-flopped! This is why all these key points seem so familiar!! The shape of the curve will always be: (a,1) (a,1) (1,0) (1,0) (1/a, -1) (1/a, -1) Where a > 1 Where 0< a < 1 Increasing Function Decreasing Function Example: On the graph below, graph both f(x) = 4x and y g(x) = (1/4)x (Also f(x) = 4-x) and h(x) = log4 x x Y. Butterworth Ch. 9 Summary 4 Remember our translations from chapter 2? Exponential functions can be translated too. The vertical translations move the asymptote from y = 0 to whatever the constant added to the functions is. This means that they translate our points being graphed by the constant “c” in the y-coordinate. f(x) = ax + D graphs with (0, 1 + D), (1, a + D) & (-1, 1/a + D) Example: Graph y f(x) = 2x and g(x) = 2x + 3 x Note: Notice how the asymptote for the graph of g(x), went from being the x-axis to the line 3 units higher, the line y = 3. Recall also that we talked about the horizontal translations of graphs in Chapter 2 as well, these are translations that move the x-coordinate. f(x) = ax – C graphs with (0 + C, 1), (1 + C, a) & (-1 + C, 1/a) Example: Graph f(x) = 2x–3 on the graph above. Note: The y-intercept is no longer the point (0, 1). It is now the point (0, a–C). In our case that is (0, 1/8) We aren’t going to cover log translations. Y. Butterworth Ch. 9 Summary 5 Solving Log & Exponential Functions First, let’s recall our inverse functions. Our 1st task will be to switch between the two functions – logs to exponentials and exponentials to logs. f(x) = ax where a > 0, but a 1 D: (-∞, ∞) R: (0, ∞) f(x) = log a x where a > 0, but a 1 D: (0, ∞) R: (-∞, ∞) We want to be able to switch seamlessly between the two inverse functions. This will help us in our equation solving skills later on. The key is that the domain of the exponential (the x) becomes the range of the logarithmic (the y), and the range of the exponential (the y) becomes the domain of the logarithmic (the x). If y = ax then loga y = x is the logarithmic equivalent. Example: 102 = 100 becomes log 100 = 2 Note: There is no base written on the log in this case because the base 10 is what we call a common log and when you don’t see a base written it is assumed to be base 10. You Try: Write the logarithmic equivalent for: 2 a) 5 = 25 b) 2(x + 1) = 8 c) 43 = z Of course we can always go the other way too. Example: log3 x = 2 becomes 32 = x You Try: Write the exponential equivalent for: a) log7 x = 3 b) loga 9 = 2 c) log 10 = z Let’s introduce another special base and continue on. The next base is the natural log which is base e. This is another irrational number like π. It too is naturally occurring in natural processes having to do with growth and decay. Like π, it too has a common approximation. e ≈ 2.7183. When the base of a logarithm is base e, we call it the natural log and it looks like: ln x = y Example: Write the equivalent form: a) ex = 1 b) ln z = 5 Y. Butterworth Ch. 9 Summary 6 Let’s discuss finding a logarithm’s value on your calculator. If we have a common log or a natural log it is a simple plug and chug calculation, but with any other base we will need something called the base change formula. Let’s get the formula and then practice the common and natural log calculations and then extend to the use of the base change formula with these calculations: loga b = log b = ln b log a ln a Note: That’s log of argument divided by log of base. Example: Find the approximate value of the log to the nearest ten-thousandth. a) log 5 b) ln 2 Example: Use the base change formula to find the exact value of the following and then approximate to the nearest ten-thousandth using your calculator. a) log2 5 b) log9 70 You Try: log7 9 Now, we want to investigate solving exponential and logarithmic equations using the a trick. The trick will be rewriting the exponential form so that the base is the same on each side of the equation. This works because of the following property (which is really common sense): If ax = ay (a is the same base), then x = y. Example: Rewrite each side so that you see like bases raised to the same power. Use the above property to find the value of the variable (you may need to solve an algebraic equation to accomplish the task). x 2 a) 5 = 5 b) 3x = 81 c) (1/16)x = 24 d) 2(x + 1) = 4x e) 9(2x + 3) = 3x Now, let’s extend this concept to the logarithmic equations. We must first write the logarithmic equations in their exponential form, and then we can solve. Example: Rewrite in exponential form and then use the above property to solve. a) log2 4 = y b) log27 3 = x c) log3 1/27 = z Y. Butterworth Ch. 9 Summary 7 You Try: Using the principle that if the bases are the same then the exponents are equal, solve each of the following. a) 9x = 1/81 b) log81 9 = x c) log 10,000 = y Now, another principle that can help us solve exponential and logarithmic equations. It again revolves around like bases, but it focuses on the logarithmic equation. If loga x = loga y (the bases are the same), then x = y (note these are called the arguments). Example: Solve the following. Make sure your solution is valid under the domain of the log function – (0, ∞). a) log (x + 1) = log 2x You Try: ln (x – 1)2 = ln (3x – 5) Two more related principles. This one is a little trickier. In this one you manipulate the equation by making the original the exponent. aloga (x) = x In other words, if the bases are the same then the argument of the logarithm in the exponent is the answer. This is useful when there is no possible way of rewriting the bases so that they are the same. Example: Solve using this principle. a) log7 x = 2 b) logx 64 = 2 You Try: log9 x = 1/2 Now, the similar principle for exponentials. It requires taking the log of both sides and again revolves around the fact that the exponential can’t be rewritten to have the same base. You may also need a calculator for this one! loga ax= x Example: Solve the following. x a) 3 = 4 b) 10x + 1 = 2 c) ex = 1 Y. Butterworth Ch. 9 Summary 8 You Try: Solve 7x = 20 Next, we will learn some properties that will be used to condense/expand logarithmic expressions, which will then in turn be used to solve more complex logarithmic equations where the variable is in more than one argument. These properties are based on the product, quotient and power rules for exponents so they should look somewhat familiar to you! Product Rule Quotient Rule Power Rule loga x + loga y = loga xy loga x – loga y = loga x y loga x = loga xy y Note: loga (x + y) ≠ loga x + loga y nor does loga (x – y) ≠ loga x loga y Example: Use the above rules to condense the following: a) log2 x + log2 (x + 2) b) logx 7 – logx 9 c) x log5 9 You Try: a) log2 x – log2 (x + 2) b) log7 x + log7 4 c) 2 log9 x Now the opposite direction. Example: Use the above rules to expand the following a) logx 2(x + 1) b) log17 5x c) log2 (4/3) You Try: a) log7 2/(x + 1) b) logy 5x c) logz 43 Altogether now… Example: Condense the following using all properties that apply. a) 2log2 x + 5log2 (x + 2) b) 2log3 x – log3 9 + log3 Example: Expand the following using all properties that apply. a) logx 2(x + 1)2 b) log17 5x(x + 2)3 c) log2 (4/3)x2(x + 1) Y. Butterworth Ch. 9 Summary 9 The reason that we learn these properties is so that we can use them to solve more complex logarithmic equations, such as the following. Example: Solve the following: log x + log (x + 3) = 1 As you can see we are pulling all the pieces together. 1) Condense the equation 2) Variable is in the argument so we can use the comparable exponential form to get to an algebraic equation that we can use to solve for the variable. 3) Make sure that your solution is within the domain of the log function. You Try: a) log3 x + log3 (2x – 3) = 2 b) log2 x + log2 (x – 7) = 3 c) log5 (x + 3) – log5 x = 2 d) 3 log x – 2 log x = 2 Y. Butterworth Ch. 9 Summary 10