Question

I need help finding the inverse of this function. g(x)= (-1/(x-1)) + 3

Answer 1

Please show your work. :)

Answer 2

you need a function f(x) such that f(x)g(x)=1 so \[f(x) \cdot \frac{ -1 }{x-1 } +3f(x)=1 \]\[-f(x)+3(x-1)f(x)=x-1\]
\[f(x)(3x-4)=(x-1)\]
\[f(x)=\frac{ x-1 }{ 3x-4 }\]

Answer 3

I know the answer is g-1(x) = (-1/(x-3)) + 1 I just do not know how to get to this answer

Answer 4

oops your'e right ,
\[g(x)= \frac{-1}{x-1} +3 \]
\[(x-1)g(x)=3x-4\]
\[(3-g(x))x=4-g(x)\]
\[x=\frac{4-g(x)}{3-g(x)}\]

replacing g(x) with x , and x with g^-1(x) :
\[g^{-1}(x)=\frac{4-x}{3-x} =\frac{x-4}{x-3}=\frac{x-3}{x-3}-\frac{1}{x-3}=-\frac{1}{x-3}+1 \]

Answer 5

the last step is because \[(g \circ g^{-1})(x)=x\]
it's not 1 of course :) too much set theory heh...

Answer 6

What would you do for this problem
\[g(x) = (-4 +\sqrt[3]{4x})\div2 \]
I know the answer is
g -1(x) = 2(x + 2)

Answer 7

*g -1(x) = 2(x + 2)^3

Answer 8

again you need to get the x "out" :
\[2g(x)+4=\sqrt[3]{4x}\]
\[x=\frac{(2g(x)+4)^{3}}{4}=\frac{2^{3}(g(x)+2)^{3}}{4}=2(g(x)+2)^{3}\]

Answer 9

it's just algebra really.

Answer 10

thanks

Answer 11

you welcome