# Trouble with a domain problem

• September 24th 2010, 06:31 PM
cb220
Trouble with a domain problem
Hi, I'm having trouble with a domain problem, I think I'm going about it the wrong way, or maybe I'm missing something.

$f(x) = \sqrt{2x+2}$
$g(x) = (x-5)/x$
$(f o g)(x) = ?$

I need to find the domain of (f o g)(x). I started by finding (f o g)(x). I got:

$(f o g)(x) = \sqrt{(4x-10)/x}$

I've found that f(x) must be greater than -1 and g(x) cannot equal 0. So the ranges so far would be:

[-1, 0)U(0,inf)

Then I looked at the limitations of (f o g)(x). I found that it x cannot equal 0, x must be greater than or equal to $\frac{5}{2}$. So as I see it, the limits I listed for f(x) and g(x) can be disregarded, since x must be greater than 2.5, which is bigger than -1 or 0.

[ $\frac{5}{2}$,inf)

Or so I thought, I tried to submit that answer and it came back as incorrect.

Am I solving this the wrong way? Thanks in advance.
• September 24th 2010, 07:11 PM
tonio
Quote:

Originally Posted by cb220
Hi, I'm having trouble with a domain problem, I think I'm going about it the wrong way, or maybe I'm missing something.

$f(x) = \sqrt{2x+2}$
$g(x) = (x-5)/x$
$(f o g)(x) = ?$

I need to find the domain of (f o g)(x). I started by finding (f o g)(x). I got:

$(f o g)(x) = \sqrt{(4x-10)/x}$

I've found that f(x) must be greater than -1 and g(x) cannot equal 0.

I suppose you meant that for the function f it must be that $x\geq -1$ , and for g it must be $x\neq 0$

So the ranges so far would be:

[-1, 0)U(0,inf)

Then I looked at the limitations of (f o g)(x). I found that it x cannot equal 0, x must be greater than or equal to $\frac{5}{2}$. So as I see it, the limits I listed for f(x) and g(x) can be disregarded, since x must be greater than 2.5, which is bigger than -1 or 0.

[ $\frac{5}{2}$,inf)

Or so I thought, I tried to submit that answer and it came back as incorrect.

Am I solving this the wrong way? Thanks in advance.

Well, since for f it must be $x\geq -1$ , we then have that for $f\circ g(x):=f(g(x))$ it must be

$g(x)=\frac{x-5}{x}\geq -1\Longleftrightarrow \frac{2x-5}{x}\geq 0\Longleftrightarrow x\leq 0\,\,or\,\,x\geq \frac{5}{2}$.

The above, together with the first limits, gives you the answer: $-1\leq x <0\,\,or\,\,x\geq \frac{5}{2}\Longleftrightarrow [-1,\,0)\cup (5/2,\,\infty)$

Tonio
• September 24th 2010, 07:20 PM
cb220
Quote:

Originally Posted by tonio
$g(x)=\frac{x-5}{x}\geq -1\Longleftrightarrow \frac{2x-5}{x}\geq 0\Longleftrightarrow x\leq 0\,\,or\,\,x\geq \frac{5}{2}$.

Thanks for the quick reply. I'm with you until $x\leq 0$. Not sure how to arrive at that, with the direction of the sign changed.

That said, the rest of your post makes sense, and I can see why my answer was wrong if I could only figure out how to get the $x\leq 0$.

EDIT: Gah, the answer you've provided also comes back as incorrect :( I've tried it as [-1,0)U(5/2,inf) and [-1,0)U[5/2,inf). Can't see why it isn't working.
• September 24th 2010, 08:19 PM
tonio
Quote:

Originally Posted by cb220
Thanks for the quick reply. I'm with you until $x\leq 0$. Not sure how to arrive at that, with the direction of the sign changed.

That said, the rest of your post makes sense, and I can see why my answer was wrong if I could only figure out how to get the $x\leq 0$.

EDIT: Gah, the answer you've provided also comes back as incorrect :( I've tried it as [-1,0)U(5/2,inf) and [-1,0)U[5/2,inf). Can't see why it isn't working.

Of course: I did an elementary mistake, but it doesn't matter. you should understand the way by which we arrive to the answer and reach it...

As before, it must be $g(x)\geq -1\Longleftrightarrow x\,\,\leq 0\,\,or\,\, x\geq \frac{5}{2}$ , and now we must take care that for g,

$x\neq 0$ , so the answer must be $x< 0\,\,or\,\, x\geq\frac{5}{2}\Longleftrightarrow (-\infty,\,0)\cup [5/2,\,\infty)$

Tonio
• September 24th 2010, 08:36 PM
cb220
Hmm, okay I think I get that. ${x}\geq{-1}$ isn't used in the final solution because that was a limiter put on f(x), but the x in f(x) isn't used in (f o g)(x), instead g(x) is used. Is that right?

On the other hand I'm still having a hard time seeing how to get:

$g(x)\geq -1\Longleftrightarrow x\,\,\leq 0$

I got the $\frac{5}{2}$ part of it but not getting the other half.
• September 24th 2010, 09:04 PM
tonio
Quote:

Originally Posted by cb220
Hmm, okay I think I get that. ${x}\geq{-1}$ isn't used in the final solution because that was a limiter put on f(x), but the x in f(x) isn't used in (f o g)(x), instead g(x) is used. Is that right?

On the other hand I'm still having a hard time seeing how to get:

$g(x)\geq -1\Longleftrightarrow x\,\,\leq 0$

I got the $\frac{5}{2}$ part of it but not getting the other half.

$\frac{x-5}{x}\geq -1\Longleftrightarrow \frac{x-5}{x}+1\geq 0\Longleftrightarrow \frac{2x-5}{x}\geq 0\Longleftrightarrow x(2x-5)\geq 0$ , and

now you've a simple quadratic inequality (draw the hiperbole of the left hand and check when it is above the x-axis...

Tonio