let g:(0,inf) when all real numbers be g(x)=x-(1/x) show that g has an inverse. I've already prove that g does not have an inverse when the domain is (-inf,0)U(0,inf) and i got the inverse of y=x-(1/x) which I got y=[x+-sqrt(x^2+4)]/2 but how do I prove this is an inverse?

owlking299 Dec 30, 2021

#1

#2**+1 **

I used algebra and got the inverse but how can i prove that it is an inverse without graphing it?

owlking299
Dec 30, 2021

#3**+2 **

Hello :))

I got the same inverse as you.

I think the way to prove that it's an inverse it to show that g(x) is a one to one function on the domain (0, infinity), basically saying that no two numbers from that domain when plugged into x - 1/x will have the same output.

Perhaps you can prove it by showing that g(x) is a strictly increasing function on that domain because if a function is strictly increasing then no two different numbers can have the same output?

I'm not too confident about that.

a is a number that is greater than 0.

f(g + a) > f(g)

g + a - 1/(g + a) > g - 1/g

g + a > g

a > 0 (true)

1/g > 1/(g + a) Since both g and a are positive numbers, g + a must also be positive.

g + a > g

a > 0 (true)

Therefore,

g + a - 1/(g + a) > g - 1/g

Since the function is strictly increasing, no two different inputs can have the same outputs since one of the numbers must be greater than the other, meaning their output must also be greater than the other.

=^._.^=

catmg Dec 31, 2021

#4**+2 **

Ohh that makes more sense. I never thought about using a more abstract algebra or proving its strictly increasing. Thank you!

owlking299
Dec 31, 2021