# Domain f o g

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• 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.

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

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

For instance

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

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?

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

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
Grandad
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.

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

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

For instance

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

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?

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

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

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

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

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

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

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

Grandad