Find h⁻¹(x): Expression for x in terms of y

[nirvana1990]

If h(x)=(3x-5)/(7-2x)

Find an expression for h^-1(x)





Here's my attempt!

y=(3x-5)/(7-2x)

(swap x for y): x=(3y-5)/(7-2y)

I've tried rearranging to find y in terms of x but I can't see how to do it!

x(-2y)=(3y-5)/7

-2y=(3y-5)/7

-2y/3y=-5/x

 

[Kurdt]

You don't swap them in that manner. You try and rearrange so you get x as a function of y. More an issue of semantics I believe.

 

[EnumaElish]

Let's say your function was y = f(x) = 1/(1+x) over x < -1 or -1 < x. Then 1/y = 1+x so x = 1/y - 1. That is x = f^-1(y) = 1/y - 1 over y < 0 or 0 < y.

 

[JonF]

Did you try to multiply both sides by the denominator, distribute the y, then group your terms with x, then factor out the x?

 

[nirvana1990]

Oh i think I've got it:

y=(3x-5)/(7-2y)
x=(3y-5)/(7-2y)
7x-2xy=3y-5
-2xy-3y=-5-7x
y(-2x-3)=-5-7x
y=(-5-7x)/(-2x-3)=h^-1(x)

 

[nirvana1990]

Thanks for your help! xxx

 

[EnumaElish]

You should start with y = (3x-5)/(7-2x) [not y=(3x-5)/(7-2y)] then apply the rules of elementary algebra until you have x = g(y).

I'll give you the first step: (7-2x)y = (3x-5)

 

[JonF]

You should start with y = (3x-5)/(7-2x) [not y=(3x-5)/(7-2y)] then apply the http://en.wikipedia.org/wiki/Elemen...lementary_algebra]rules of elementary algebra until you have x = g(y).

I'll give you the first step: (7-2x)y = (3x-5)

Some books teach the variable switch first… but normally you show it in one step, not 2.

p.s. you probably want to factor out those negatives just to make it look pretty.

 

[EnumaElish]

Oh i think I've got it:

y=(3x-5)/(7-2y)
x=(3y-5)/(7-2y)
7x-2xy=3y-5
-2xy-3y=-5-7x
y(-2x-3)=-5-7x
y=(-5-7x)/(-2x-3)=h^-1(x)

This is correct, except I would've switched the variables at the end.

 

[Kurdt]

Thats interesting, I've never seen the variables being swapped first so that threw me a bit. Is that just to maintain the notation that we usually have y = f(x).

 

[JonF]

Thats interesting, I've never seen the variables being swapped first so that threw me a bit. Is that just to maintain the notation that we usually have y = f(x).

no clue, i always thought it was silly myself. But they do.

 

[Dick]

Thats interesting, I've never seen the variables being swapped first so that threw me a bit. Is that just to maintain the notation that we usually have y = f(x).

Yes, it is. If the initial function is given as y=f(x) the inverse function should also probably be stated as y=f^(-1)(x) so x is the independent variable in both. That's all. You can swap at the end if you like.