Is the Inverse a Function?Date: 9/2/96 at 16:16:17 From: Ralph Senter Subject: Compositions and Inverses of Functions I am lost on how to do these problems. Can you help me? Find the inverse of each function. Then state whether the inverse is a function. 1. f(x) = 4x+4 2. f(x) = x(cubed) 3. f(x) = x(squared)-9 Date: 9/3/96 at 3:32:47 From: Doctor Mike Subject: Re: Compositions and Inverses of Functions Hello Ralph, 1. If a function f takes x to some expression in x, then the inverse has to "undo" what was done by f. In this case it's pretty easy. f starts with x, multiplies it by 4, and then adds 4. To reverse that process you subtract 4, and then divide by 4. Often the inverse of a function f is written as f with a "-1" superscript, but for typing here I'm going to use "i" for it. Then we have : x - 4 i(x) = ------- 4 As a final check, it must be true that i(f(x))=x , that is, if you start with x and use a function f to get the f(x) result, and then use the function i on that to get a second result i(f(x)), then this second result must be just x again. Let's see: f(x) - 4 (4x+4) - 4 4x i(f(x)) = ---------- = ------------ = ---- = x 4 4 4 The inverse in this example is a function, because for every x, the expression defining i(x) can have only one value or meaning. For contrast see number 3. 2. Hint: To reverse the cubing process, what do you do? That is, how do you get from x(cubed) back to x? 3. f starts with x, squares it, and subtracts 9. To reverse that process you add 9 back on, and then take the square root. The formula for this inverse function "i" is i(x) = sqrt(x+9) where sqrt represents the square root. The trouble is, a number can have both a positive square root and a negative squart root, so i(7) = sqrt(16) could be either +4 or -4. This means the inverse in this example is not a function. Such a rule for starting with a particular x and getting non-unique results is often called a relation, because it specifies a relationship which does not fully qualify as a functional relationship. I hope this helps you to get a handle on these concepts. -Doctor Mike, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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