# Another question about the domain in a compound function

1. Jun 24, 2007

### 0000

¿Could the domain of a compound function be obtained in the same way that non-compound functions?

I think the answer is not, like in this example:

f(x)=1/x²
g(x)=√(2x-6)

f(g(x))=1/(√(2x-6))²

Recently Hurkyl explained me that (a^b)^c is not always equal to a^bc, although, I've proved this step with Derive and it seems to be fine.

f(g(x))=1/(2x-6)

Back to the matter, the domain of the function H(x)=1/(2x-6) is R - {3}

But the domain of f(g(x))=1/(2x-6) is (3,∞)

So, Am I right in the fact that a compound function can have a different domain than the "same function" which isn't the product of a composition of functions?

As always, excuse me if my english isn't very clear.

Last edited: Jun 25, 2007
2. Jun 25, 2007

### HallsofIvy

Staff Emeritus
Note that a function has two parts- a domain (possible values of x) and a rule connecting each x with a unique a. That is "f(x)= x2 for x> 0" is not the same function as "f(x)= x2". In the formula
$$f(g(x))= \frac{1}{(\sqrt{2x-6})^2}$$
since the domain of g is$x\ge 3$, the square root cannot be taken for x< 3 and so the domain is $x\ge 3$.

$$f(g(x))= \frac{1}{(\sqrt{2x-6})^2}$$
is NOT the same function as
$$\frac{1}{2x-6}$$.

This is the reason Calculus texts refer to the "deleted" neighborhood of a when taking limits at a. In order to take $\lim_{x\rightarrow 2} (x^2-4)/(x-2)$ you must note that $(x^2-4)/(x-2)$ for all x except x= 2 and use the theorem "If f(x)= g(x) in some deleted neighborhood of a, then they have the same limit at x= a".

3. Jun 25, 2007

### 0000

Thanks, I think I see the things a little more clear now, but why when I simplify
1/(√(2x-6))² with Derive I get 1/2(x-3)?

4. Jun 25, 2007

### VietDao29

Err... What's Derive? Some computer programme, right? Btw, just don't rely on them so much. :)

Ok, here it goes.

In fact, the statement:
$$\left( \sqrt{x} \right) ^ 2 = x$$ is incorrect.

When saying f(x) = g(x), it must be true that:
1. f(x), and g(x) have the same domain.
2. They (f(x), and g(x)) return the same value for every x in their domain.

In this case:
$$\left( \sqrt{x} \right) ^ 2 = x$$ violates the first requirement.

The LHS (left-hand side) function is only defined for x >= 0, while the RHS (right-hand side) is defined for all x in the reals.

But, however, if you restrict the domain for the RHS function to be $$[0 ; \ \infty [$$, then they can be said to be equal, i.e: $$\left( \sqrt{x} \right) ^ 2 = x$$

--------------------

So, say, we let $$f(x) = \sqrt (x) ; \quad g(x) = x ^ 2 ; \quad h(x) = x$$, then:
f(g(x)) does not have the same domain as h(x).

Can you get it? :)

Last edited: Jun 25, 2007
5. Jun 25, 2007

### 0000

Ok, thanks, your help has been very useful. And yes, Derive it's a math program, maybe it gives that result because it works with complex numbers and in that case I think that 1/(√(2x-6))² could be equal to 1/2(x-3).

6. Jun 26, 2007

### morson

VietDao made the point that "Derive" is correct iff x ≠ 3. I'd also like to point out that if $$(\sqrt{x})^2$$ is a complex valued function, then $$(\sqrt{x})^2$$ = x for all x £ C.

In general, if f(z) is a complex valued function and f(z) = g(z) is a relation, then f(z) = g(z) for all z £ C.

If f(x) is a real valued function and f(x) = g(x) is a relation, then f(x) = g(x) for x £ D, where D is the real domain.

Last edited: Jun 26, 2007