# Math Help - Hitting bit of a snag on domain and range of functions.

1. ## Hitting bit of a snag on domain and range of functions.

Hi again all,

I am asked to find domain and range of an f(x) and the range of its inverse f^-1(x):

f(x) = sqrt(2x - 5)

So I found the domain of this to be x >or= 5/2. I'm not too good at getting range so I decided to find the domain of the inverse which would then be the range of the original. My inverse turned out to be:

f^-1(x) = (x^2 + 5)/ 2

So then it seems like the domain to the above inverse function would be all real numbers, but then I had trouble verifying that the range of my original was all real numbers cause how could the range of f(x) = sqrt(2x - 5) be a negative number?

2. Originally Posted by DannyMath
So then it seems like the domain to the above inverse function would be all real numbers, but then I had trouble verifying that the range of my original was all real numbers cause how could the range of f(x) = sqrt(2x - 5) be a negative number?
I think you are on the right track.

Given your original function is $f(x)=\pm\sqrt{2x-5}$ it will have $y \in\mathbb{R}$

But this is not a one to one function so its inverse does not exist.

Restricting the domain to force this function to be one to one you get.

Firstly

$f(x)=+\sqrt{2x-5},~y\in \mathbb{R}^- \cup \{0\}$

and then

$f(x)=-\sqrt{2x-5},~y\in \mathbb{R}^+ \cup \{0\}$

The best way to see this is via a sketch.

3. Originally Posted by pickslides
I think you are on the right track.

Given your original function is $f(x)=\pm\sqrt{2x-5}$ it will have $y \in\mathbb{R}$

But this is not a one to one function so its inverse does not exist.

Restricting the domain to force this function to be one to one you get.

Firstly

$f(x)=+\sqrt{2x-5},~y\in \mathbb{R}^- \cup \{0\}$

and then

$f(x)=-\sqrt{2x-5},~y\in \mathbb{R}^+ \cup \{0\}$

The best way to see this is via a sketch.
Ok well the original function I was given was indeed the:

$f(x)=+\sqrt{2x-5}$

function. But the problem I had was because I found that the inverse of this function had a domain of all real numbers, which made me conclude that the range of:

$f(x)=+\sqrt{2x-5}$

HAS to be all real numbers, but this doesn't seem to be the case, which is what confused me.