Inverse of a Function

[beccaw]

Hi! so Im having a lot of trouble finding the inverse of this function: y=

4

x

The given domain and range are both (0, infinity) so I already know that is what the new domain and rang will be, in i know that i have to switch the y and the x, but I'm confused about how to go about doing it. Where should I start? Should I multiply both sides by the denominator, or should I rationalize the denominator? Or should I start of be dividing by 4?

 

[srmichael]

Hi! so Im having a lot of trouble finding the inverse of this function: y=

4

x

The given domain and range are both (0, infinity) so I already know that is what the new domain and rang will be, in i know that i have to switch the y and the x, but I'm confused about how to go about doing it. Where should I start? Should I multiply both sides by the denominator, or should I rationalize the denominator? Or should I start of be dividing by 4?

The beauty of math is that you can get the answer in usually more than one way. Any of those steps you mention would be a start you can take. Though, personally, I would start in a totally different way. Square roots are never fun, so I would first square both sides. You would then have:

\(\displaystyle x^2=\frac{16}{y}\) assuming you have already made the x and y switch like you said you did.

Now, can you solve for y? Remember to put the domain restrictions on the inverse you get.

 

[beccaw]

I didn't even think of that! That makes things so much easier. Thank you!

 

[mmm4444bot]

:!: Kudos to Sir Michael for deciphering y=

4

x

 

[HallsofIvy]

:!: Kudos to Sir Michael for deciphering y=

4

x

I wondered what the check mark meant!

 

[srmichael]

I wondered what the check mark meant!

I didn't know either until I read the phrase "rationalize the denominator" by the OP