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-   -   About domain in a compound function (http://www.physicsforums.com/showthread.php?t=174813)

0000 Jun22-07 11:04 PM

About domain in a compound function
 
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.

Hurkyl Jun22-07 11:07 PM

Quote:

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

f(g(x)) = x^(1/2)
This is wrong; those two expressions are not equal.

ice109 Jun22-07 11:19 PM

Quote:

Quote by Hurkyl (Post 1362846)
This is wrong; those two expressions are not equal.

and why is that?

Hurkyl Jun22-07 11:26 PM

For example, their domains of definition are different.


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

ice109 Jun22-07 11:36 PM

Quote:

Quote by Hurkyl (Post 1362854)
For example, their domains of definition are different.


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

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

Hurkyl Jun22-07 11:55 PM

One form is
For real numbers a, b, and c: if a > 0 then [itex](a^b)^c = a^{bc}[/itex]
In fact, if you're just using plain real-number exponentiation, [itex]a^b[/itex] 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: [itex](a^b)^c = a^{bc}[/itex]
There are some other cases you can write down -- but the point is that they are all qualified.

0000 Jun23-07 12:02 PM

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.

Hurkyl Jun23-07 12:04 PM

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.


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