# About domain in a compound function

1. Jun 22, 2007

### 0000

In my book and in other places, they give this rule to obtain the domain for
a compound function: "the domain of (f o g) (x) is the set of all real
numbers x such that g(x) is in the domain of f (x)."

Then, if f(x)=x^(1/4)

and

g(x)=x^2

f(g(x)) = (x^2)^(1/4)

f(g(x)) = x^(1/2)

And applying the rule for the domain, it'll be all the real numbers. Isn't it illogical?

Thanks for the help and excuse me if there is any grammar mistakes, it's because english isn't my native language.

Last edited: Jun 22, 2007
2. Jun 22, 2007

### Hurkyl

Staff Emeritus
This is wrong; those two expressions are not equal.

3. Jun 22, 2007

### ice109

and why is that?

4. Jun 22, 2007

### Hurkyl

Staff Emeritus
For example, their domains of definition are different.

(hint: $(a^b)^c = a^{bc}$ is invalid, even though it closely resembles an identity you learned in your algebra classes)

5. Jun 22, 2007

### ice109

ok the domains difference granted, what is the actual identity then?

6. Jun 22, 2007

### Hurkyl

Staff Emeritus
One form is

For real numbers a, b, and c: if a > 0 then $(a^b)^c = a^{bc}$​

In fact, if you're just using plain real-number exponentiation, $a^b$ is only defined for a > 0. But usually we use a generalization that allows other special cases, such as integer exponents, and we have

For a nonzero real number a and integers b and c: $(a^b)^c = a^{bc}$​

There are some other cases you can write down -- but the point is that they are all qualified.

7. Jun 23, 2007

### 0000

thanks

thanks, I think it's my school teacher's fault :P, for teaching me
(a^b)^c = a^bc without explaining the constraints of that identity.

Last edited: Jun 23, 2007
8. Jun 23, 2007

### Hurkyl

Staff Emeritus
You may have a recollection of an identity such as (a^2)^(1/2) = |a| -- that one is valid for all real numbers a. This problem is very similar.