Question

find inverse

Answer 1

f(x)= 8/ (x-3)^2

Answer 2

solve
\[x=\frac{8}{(y-3)^2}\] for \(y\) in a few steps

Answer 3

\[(y-3)^2=\frac{8}{x}\]
\[y-3=\pm\sqrt{\frac{8}{x}}\]
\[y=3\pm\sqrt{\frac{8}{x}}\]

Answer 4

How about this one ?
(x-1)^3 / 8

Answer 5

@zepdrix plz help

Answer 6

find inverse @zepdrix

Answer 7

i did it but my answer came out different form the book :(

Answer 8

\[\large y=\frac{(x-1)^3}{8}\]So we start by swapping the x and y,\[\large x=\frac{(y-1)^3}{8}\]Let's multiply both sides by 8,\[\large 8x=(y-1)^3\]Take the cube root of both sides,\[\large \sqrt[3]{8x}=y-1\]Add 1 to each side,\[\large y=\sqrt[3]{8x}+1\]8 is a perfect square, let's go ahead and pull that out of the root. And I'm also going to write the x using a rational expression, it just looks nicer that way :)\[\large f^{-1}(x)=8x^{1/3}+1\]

Answer 9

Luckily we don't need to worry about adding a \(\large \pm\) to this root, since it's an `odd` root.

Answer 10

Woops, sorry i pulled the 8 out and I didn't take the root LOL

Answer 11

oh cube root.... i see

Answer 12

\[\large f^{-1}(x)=2x^{1/3}+1\]
The cube root of 8 is 2, sorry lil typo there :3

Answer 13

Yah you had a 3rd power, so you need to take the 3rd root in order to clean that up :)

Answer 14

on the back of the book, it just took the 2 from the 8
2√x +1

Answer 15

i don't know how to do that cube root so...lol

Answer 16

hmm