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. 
Quote:

Quote:

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) 
Quote:

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 realnumber 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. 
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. 
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.

All times are GMT 5. The time now is 06:26 AM. 
Powered by vBulletin Copyright ©2000  2014, Jelsoft Enterprises Ltd.
© 2014 Physics Forums