# Domain f o g

• Sep 7th 2009, 04:45 AM
Peleus
Domain f o g
Hi all,

First things first, what I've seen as the answer for these is described as follows.

$
Dom (f o g) = {x : x \in Domg-and-g(x) \in Dom f}
$

I'm not sure what is meant by the second section though, g(x) in Dom f?

For instance

$f(x) = \sqrt{4 - x^2}$
$g(x) = \ln{x}$

So the dom for f(x) is [-2, 2]
the dom for g(x) is (0, inf]

How does it work fitting into this formula?

$Dom(fog) = x : x \in x > 0 and \ln{x} \in [-2, 2]$

So Dom(fog) = 0 < x < 2 ???

I know this is incorrect, but this is how my logic of applying the formula works.

Anyone who can point out my mistakes or explain it in english a bit better would be appreciated.
• Sep 7th 2009, 05:36 AM
Domain of composite function
Hello Peleus
Quote:

Originally Posted by Peleus
Hi all,

First things first, what I've seen as the answer for these is described as follows.

$
Dom (f o g) = {x : x \in Domg-and-g(x) \in Dom f}
$

I'm not sure what is meant by the second section though, g(x) in Dom f?

For instance

$f(x) = \sqrt{4 - x^2}$
$g(x) = \ln{x}$

So the dom for f(x) is [-2, 2]
the dom for g(x) is (0, inf]

How does it work fitting into this formula?

$Dom(fog) = x : x \in x > 0 and \ln{x} \in [-2, 2]$

So Dom(fog) = 0 < x < 2 ???

I know this is incorrect, but this is how my logic of applying the formula works.

Anyone who can point out my mistakes or explain it in english a bit better would be appreciated.

As I'm sure you know, when we 'feed' the value $x$ through the composite function $f \circ g$, $x$ is first 'fed into' $g$, and then the output of $g$ (which is represented by $g(x)$) is then used as the input to the function $f$. To ensure a valid input into $f$, then, we must make sure that the output $g(x)$ lies in the domain of $f$. That is what is meant by $g(x) \in Dom(f)$.

As far as the example you quote is concerned, $g(x) = \ln(x)$, any value of $x$ in the domain $(0, +\infty)$ will produce an output from $g$, but not every output can then be used as an input to $f$: only those outputs that are themselves in $[-2,2]$.

So you need to ask yourself the question: What values of $x$ can I use as inputs for $\ln(x)$ that will produce outputs in the range $[-2,2]$ ? The answer will be given by the solutions to the equations $\ln(x_{min}) = -2$ and $\ln(x_{max}) = 2$.

Raising $e$ to the power of both sides of each equation we get:

$x_{min} = e^{-2}$ and $x_{max} = e^2$

There's the domain of $f\circ g$, then: $[e^{-2},e^2]$