The Student Room Group
Reply 1
Alevelstudent123
Hi,

I was wondering if you find the inverse of y=2/x-1, wouldn't it be one of either:

2/x+1=y

or

2-x/x=y?

The top one is apparently wrong, I don't see how though. To me it is just another method with an alternative outcome!

Could anyone help me out? I'd really appreciate it!


Your question is ambiguous.

Do you mean y=2x1 y = \frac{2}{x} -1 or y=2x1 y = \frac{2}{x-1} ?
Alevelstudent123
Hi,

I was wondering if you find the inverse of y=2/x-1, wouldn't it be one of either:

2/x+1=y

or

2-x/x=y?

The top one is apparently wrong, I don't see how though. To me it is just another method with an alternative outcome!

Could anyone help me out? I'd really appreciate it!

I'm going to assume that your equation is y=(2/x)-1, but it would be useful if you could make sure that you use brackets in future posts to avoid ambiguity. In this case, the inverse should be y=2/(x+1), as you suggest.

However, if you meant y=2/(x-1) as the original equation, then this has a different inverse. :smile:
if y=2/(x+1) Inverse (x/2)-1

if y=(2/x)+1 Inverse (x/2)-1/2
victor_b1992
if y=2/(x+1) then dy/dx 2/x +1

if y=(2/x)+1 then dy/dx 2x + 2

Have you read what the OP is actually asking for? Differentiation doesn't come into their question, and you've actually given incorrect derivatives anyway. :no:
y=2/(x-1)
y(x-1)=2
yx-y=2
x=2+y/(y)
f^-1(x)=2+x/(x)
?
Reply 6
If the question should read y = (2/x)-1, then your answer is correct. If you mean 2/(x-1), then the answer would be different.

EDIT: See the above post.
Illusionary
Have you read what the OP is actually asking for? Differentiation doesn't come into their question, and you've actually given incorrect derivatives anyway. :no:


sry i mean inverse :smile:
victor_b1992
sry i mean inverse :smile:

...but they're not correct as inverses, either.
Illusionary
...but they're not correct as inverses, either.

ok well it right nw lol just didn't no where my head was at this question lol.:smile:
Reply 10
Assuming your equation is y=2x1y = \frac{2}{x-1}

Cross multiply, so that:

y(x - 1) = 2

yx - y = 2

yx = 2 + y

x=2+yyx = \frac{2+y}{y}

Re-writing:

y=2+xxy = \frac{2+x}{x}

So:

f(x)1=2+xxf(x)^{-1}= \frac{2+x}{x}
As a notational point (call me a pedant), this is f1(x)f^{-1}(x), not f(x)1f(x)^{-1}, which is completely different.