Question

How to find the inverse of 1 over (x+2)^2

Answer 1

\[y = \frac{ 1 }{ \sqrt{x} }-2\]

Answer 2

how did you get that?

Answer 3

do you want me to show you how?

Answer 4

sure

Answer 5

yes please

Answer 6

step one: \[x = \frac{ 1 }{(y+2)^2 }\]

Answer 7

step two: take the sqrt of both sides
\[\sqrt{x}(y+2)=1\]

Answer 8

did get this step?

Answer 9

\[y+2 = \frac{ 1 }{ \sqrt{x} }\]

Answer 10

oh ok. so to get rid of the square, you root it

Answer 11

\[y = \frac{ 1 }{ \sqrt{x} }-2\]

Answer 12

\[f^{-1}(x)=\frac{ 1 }{ \sqrt{x}}-2\]

Answer 13

I hope this help! ask me if you are having a problem with any step...

Answer 14

when yo took the root, y did you have the y+2 beside it. Bc if I do it, I will have root x on one side and then 1 over y+2 on the other.

Answer 15

yes you can do that, i just combined two steps in one

Answer 16

ok, so after that what would I do? Would I have to multiply both sides by y+2

Answer 17

yes

Answer 18

then I would divide by the root x and subtract2 from both sides

Answer 19

exactly well done

Answer 20

did you notice that i swapped the x and y in the first step

Answer 21

yes. I remember being told to do that when I was taught about inverses. Could I not swap them and have the inverse = x

Answer 22

it is easier to swap, otherwise more likely to make an error...

Answer 23

oh ok. Thank you

Answer 24

you are welcome! all the best...

Answer 25

That's only half of the inverse relation, however. You didn't take the negative root into account.

\[f(x) = y = \frac{1}{(x+2)^2}\]Swap variables
\[x = \frac{1}{(y+2)^2}\]
Take reciprocal of both sides\[\frac{1}x = (y+2)^2\]Take the square root of both sides\[\pm\frac{1}{\sqrt{x}} = (y+2)\]Subtract 2 from both sides\[f^{-1}(x)= -2\pm\frac{1}{\sqrt{x}}\]

In the attached graph, you can see what happens if you only use one root: you only get the green or red portion, not both.