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### Is the Inverse a Function?

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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/
```
Associated Topics:
High School Basic Algebra
High School Functions

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