If a function has an inverse then we can determine the input if we know the output. For example if the function

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1 1 Inverse Functions We know what it means for a relation to be a function. Every input maps to only one output, it passes the vertical line test. But not every function has an inverse. A function has no inverse if there are outputs that are mapped to by different inputs such as in the case of y = x 2. If a function has an inverse then we can determine the input if we know the output. For example if the function f(x) =x 3 7 gave an output of 20 what was the input? What about if the output was 5.167? How are you solving this? The function you are using is the inverse, f 1 (x). Find it algebraically. Graphically, a function has an inverse if it passes the horizontal line test. A function that has an inverseissaidtobeaone to one function. That means every one input goes to only one output. In math we can write x 1 6= x 2 f(x 1 ) 6= f(x 2 ) Prove the example above is a one to one function. Begin by assuming that x 1 6= x 2 but f(x 1 )=f(x 2 )

2 and prove by contradiction. Of course that makes good sense because we already found the inverse and we can draw the graph and show it passes the horizontal line test. So let s say a function, y = f(x) is one-to-one with domain A and range B. The inverse function, f 1 (y) =x, has domain B and range A. Notice how the range and domain switch places. So what s f 1 (f(a)) = f(f 1 (b)) = and where does the a and b come from? So we know how to algebraically get f 1 (x) from f(x) but how do we get the graph of the inverse from the original graph? As an example of the technique, sketch the graph of y = e x and then find its inverse.

3 2 Logarithms The logarithm function, y =log b (x) is defined for all b>0,b6= 1, where b is the base. The graph looks like x for all b s. We can see that the range is all the reals and the domain is (0, ). This introduces a new complication into our find the domain of the function questions. Before this we worried about not dividing by zero and not taking the square root of a negative number. Wealsohavetomakesurenottotakethe log of zero or a negative number. So what s logarithm? Well, when I write log b (x) I am saying What do I have to raise b to, to get x?. In other words y =log b (x) is equivalent to b y = x Ifthereisnobasewritten,itisassumedtobe base 10, like on your calculator. That is, log(100) = 2.

4 Q? Evaluate log 2 (16). A. What number do I have to raise 2 to to get 16?, Well 4, since 2 4 =16 Q? Evaluate log 3 ( 1 27 ). A. What number do I have to raise 3 to to get 1 27?, Well it must be negative to make the fraction 1 over and note that 3 3 =27, therefore log 3 ( 1 27 )= 3. That is 3 3 =27. Q? Evaluate log 29 (29). A. What number do I have to raise 29 to to get 29?, Well 1, since 29 1 =29. This would have been the answer for any base, not just 29. Q? Evaluate log 17 (1). A. What number do I have to raise 17 to to get 1?, Well 0, since 17 0 =1. Thiswouldhavebeen the answer for any base, not just 17.

5 2.1 The Natural Logarithm, ln(x) The logarithmic function y =log e (x) occurs so often that it was given its own symbol, ln(x). That is ln(x) =log e (x). The rules for regular log s apply for ln(x) also. The function y =ln(x) is the inverse function to y = e x, that is, they undo each other. ln(e x )=x e ln(x) = x We will use this fact to solve for x. Try taking the ln(e 17 ) on your calculator. Your calculator probably only has log base 10, log(x), and base e, ln(x). Q? Solve for x, e 2x 2 =7 A. Take the ln() of both sides. e 2x 2 =7 ln(e 2x 2 )=ln(7) 2x 2=ln(7) 2x =ln(7)+2 x = ln(7) Note that ln(7) can be approximated on a calculator, it s just a decimal number and is treated as such in the above algebra.

6 Q? Solve for x, ln(x 7) = 29 Q? Find the domain of the function y = 1 ln(x 2) A. There is no root sign. In order not to divide by zero, ln(x 2) 6= 0 e ln(x 2) 6= e 0 x 2 6= 1 x 6= 3 Now we just have to worry about only taking the ln() of a strictly positive number, that is, x 2 > 0,x > 2. Therefore the domain of the function is (2, 3) (3, ). 2.2 Change of Base Formula Ok, if my calculator only has log base 10 and base e, how do I evaluate something like log 3 (4)? We change it into base 10 or base e. log b (x) = log a(x) log a (b) log b (x) = ln(x) ln(b) = log(x) log(b) Because we get to choose a in the above formula we can make a = e or a =10to get the second

7 equations. So log 3 (4) = ln(4) ln(3) ' The Logarithm Rules log b (AC) =log b (A)+log b (C) log b ( A C )=log b(a) log b (C) log b (A) x = x log b (A) log b ( 1 C )= log b(c) All the above rules are valid for ln(x) also. We can use them to solve for x andtosimplify equations. Before we do some examples, let me say what these rule do not say. It is not true that log b (A C) = log b(a) log b (C) or log b(a + C) =log b (A)log b (C). These two misconceptions are common. Don t fall for it. Q? Solve for x, ln(x)+ln(x +1)=ln(2) A rule of thumb for solving for x is: If the x is inside a ln() use the exponential function to get it out, if the x is inside an exponential function, take the ln() to get it out.

8 Section 1.6, # 33, 35-40, 42, 47, Submit 38, 42, 50 Friday.