Question

Need help with inverses

Answer 1

\[({x^4 \over {1-x^4}})^{-1} = {1-x^4 \over x^4} = {1 \over x^4} - 1\]Correct?

Answer 2

\(\bf \left(\cfrac{x^4}{1-x^4}\right)^{-1}\quad ?\)

Answer 3

are you supposed to ... find the inverse of that?

Answer 4

Its part of a much larger problem but ya. I just want to make usre I'm doing the inverse correctly before I move on

Answer 5

I think my first post is correct but I think I might have broken a rule with the second post

Answer 6

... one sec

Answer 7

You're trying to find the `reciprocal` of the fraction?
Your first post looks good.

I'm not sure what you're doing in the second one... hmm

Answer 8

\(\bf {\color{red}{ y}}=\left(\cfrac{{\color{blue}{ x}}^4}{1-{\color{blue}{ x}}^4}\right)^{-1}\qquad inverse\implies {\color{blue}{ x}}=\left(\cfrac{{\color{red}{ y}}^4}{1-{\color{red}{ y}}^4}\right)^{-1}\\ \quad \\ \quad \\
x=\cfrac{1-y^4}{y^4}\implies x=\cfrac{1}{y^4}-\cfrac{y^4}{y^4}\implies x=y^{-4}-1\\ \quad \\
x+1=\cfrac{1}{y^4}\implies y^4=\cfrac{1}{x+1}\implies \large y=\sqrt[4]{\cfrac{1}{x+1}}\)

Answer 9

\[({y \over z}+1)^{-1}= {z \over y}+1\]is this correct

Answer 10

yes

Answer 11

\[\Large\bf\sf \left(\frac{y}{z}+1\right)^{-1}\quad=\quad \frac{1}{\left(\dfrac{y}{z}+1\right)}\quad=\quad \frac{z}{y+z}\]

Answer 12

hmmm oh yeah.... hmm .... right, can't distribute the denominator per se

Answer 13

gotcha, thank!. I thought I was breaking a rule