Question

The one-to-one function is defined by
g(x)=(2x-5)/(6+7x)
.
Find g^-1 , the inverse of g. Then, give the domain and range of g^-1 using interval notation

Answer 1

ho ho

Answer 2

pain in the butt algebra but you can do it write
\[y=\frac{2x-5}{6+7x}\] then either solve for x or switch the x and y and solve for y. either way

Answer 3

I like to use the property of inverses to solve them:
\[g(g^{-1}(x)) = x\]\[\implies x = {2(g^{-1}(x)) - 5 \over 6 + 7(g^{-1}(x))}\]

Answer 4

then solve for \(g^{-1}(x)\)

Answer 5

let \(y=\frac{2x-5}{6+7x}\). Solving for x gives \(5y+7xy=2x-5 \implies x(2-7y)=5y+5 \implies x=\frac{5y+5}{2-7y}\). So, \(g^{-1}(x)=\frac{5x+5}{2-7x}\). I hope there is no mistake.

Answer 6

Striction: \(\frac{-6}{7}\) is not in the range of \(g^{-1}(x)\) and \(\frac{2}{7}\) is not in its domain.

Answer 7

\[x=\frac{2y-5}{6+7y}\]
\[(6+7y)x=2y-5\]
\[6x+7xy=2y-5\]
\[6x+5=2y-7xy\]
\[6x+5=y(2-7x)\]
\[\frac{6x+5}{2-7x}=y\] assuming i made no algebra mistake

Answer 8

anwar i think you changed the 6 to a 5 by mistake. it is hard to write algebra here

Answer 9

I made a slight mistake. Go with satellite!

Answer 10

Yeah :)

Answer 11

i prefer to think of them as 'typos'