# Find Inverse Function

• Jul 17th 2009, 09:55 AM
magentarita
Find Inverse Function
Given f(x) = (1/x + 2) + 3, find the inverse function.

MY WORK:

Let y = f(x).

y = (1/x + 2) + 3

Switch x and y.

x = (1/y + 2) + 3

Multiply both sides by (y+2).

I ended up with

xy + 2x = 3y + 7

I then subtracted 3y and 2x from both sides.

xy - 3y = -2x + 7

I factored out a y from the left side.

y(x - 3) = -2x + 7

I finally divided both sides by (x - 3).

y = (-2x + 7)/(x - 3)

Then y becomes f^(-1) x

f^(-1) x = (-2x + 7)/(x - 3)

Is the right?

• Jul 17th 2009, 10:55 AM
Amer
Quote:

Originally Posted by magentarita
Given f(x) = (1/x + 2) + 3, find the inverse function.

MY WORK:

Let y = f(x).

y = (1/x + 2) + 3

Switch x and y.

x = (1/y + 2) + 3

Multiply both sides by (y+2).

I ended up with

xy + 2x = 3y + 7

I then subtracted 3y and 2x from both sides.

xy - 3y = -2x + 7

I factored out a y from the left side.

y(x - 3) = -2x + 7

I finally divided both sides by (x - 3).

y = (-2x + 7)/(x - 3)

Then y becomes f^(-1) x

f^(-1) x = (-2x + 7)/(x - 3)

Is the right?

it is correct
• Jul 17th 2009, 01:05 PM
magentarita
Good...
Quote:

Originally Posted by Amer
it is correct

I placed this in another forum and was told my answer is wrong but I doubted them big time.
• Jul 17th 2009, 01:21 PM
Amer
Quote:

Originally Posted by magentarita
I placed this in another forum and was told my answer is wrong but I doubted them big time.

but $x\ne 3$
• Jul 17th 2009, 02:49 PM
magentarita
Yes...
Quote:

Originally Posted by Amer
but $x\ne 3$

Yes, x cannot = 3 because it creates division by zero.
• Jul 17th 2009, 04:01 PM
yeongil
Quote:

Given f(x) = (1/x + 2) + 3, find the inverse function.
If you're not going to type in LaTex, please watch your parentheses next time! I interpreted the problem to be
$f(x) = \left(\frac{1}{x} + 2\right) + 3$
and was in the midst of writing a lengthy post correcting your work before I realized that you meant
$f(x) = \frac{1}{x + 2} + 3$.
Without LaTex, you should have written it as f(x) = 1/(x + 2) + 3.

01
• Jul 17th 2009, 04:09 PM
magentarita
ok...
Quote:

Originally Posted by yeongil
If you're not going to type in LaTex, please watch your parentheses next time! I interpreted the problem to be
$f(x) = \left(\frac{1}{x} + 2\right) + 3$
and was in the midst of writing a lengthy post correcting your work before I realized that you meant
$f(x) = \frac{1}{x + 2} + 3$.
Without LaTex, you should have written it as f(x) = 1/(x + 2) + 3.

01

Thank you so much. I now see what went wrong.