|


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/
|
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]


Ask Dr. MathTM
© 1994-2008 The Math Forum
http://mathforum.org/dr.math/