Double Division?

**JonathanEyoon** · August 30th 2007, 11:09 AM

Find f(x) / g(x)

F(x) = 2 / x

G(x) = 4 / (x + 4)

(2/x) / (4/x+4)

I kind of forgot how to tackle these kinds of problems

Oops nevermind I figured it out. Reciprocal and multiply through. Ok more important question, the result of the problem will be (x + 8) / (2x) . According the back of the book, the domain is x cannot equal 0 and -4. The thing is, clearly -4 can be used in the domain seeing as it will result in -8 in the denominator which is ok. Can someone clear this up?

**Krizalid** · August 30th 2007, 11:14 AM

$\frac{\dfrac2x}{\dfrac4x+4}=\frac{x\cdot\dfrac2x}{ x\cdot\left(\dfrac4x+4\right)}=\frac2{4+4x}=\frac1 {2+2x}$

Of course, $\forall x\ne0,-1$

**JonathanEyoon** · August 30th 2007, 11:30 AM

Originally Posted by Krizalid

$\frac{\dfrac2x}{\dfrac4x+4}=\frac{x\cdot\dfrac2x}{ x\cdot\left(\dfrac4x+4\right)}=\frac2{4+4x}=\frac1 {2+2x}$

Of course, $\forall x\ne0,-1$

I got (x + 8) / (2x) for the final answer which is the same as the answer in the back of the book

**Krizalid** · August 30th 2007, 11:38 AM

Hey, but you changed the question!

$\frac{{f(x)}} {{g(x)}} = \frac{{\dfrac{2} {x}}} {{\dfrac{4} {{x + 4}}}} = \frac{{2\left( {x + 4} \right)}} {{4x}} = \frac{{x + 4}} {{2x}}$

Is it clear now?

**JonathanEyoon** · August 30th 2007, 11:54 AM

Originally Posted by Krizalid

Hey, but you changed the question!

$\frac{{f(x)}} {{g(x)}} = \frac{{\dfrac{2} {x}}} {{\dfrac{4} {{x + 4}}}} = \frac{{2\left( {x + 4} \right)}} {{4x}} = \frac{{x + 4}} {{2x}}$

Is it clear now?

Thanks!!! I just checked with a friend's answer and he got the same one you did. Book was wrong

. Think you could tell me about why the domain for (x+4) / (2x) is X cannot = 0 and -4? If you plug in -4 to the equation, you don't get an undefined in the denominator so i'm a bit confused as to why -4. Thanks

**Krizalid** · August 30th 2007, 11:58 AM

$f(x)=\frac2x\implies\text{Dom}\,f=x\in\mathbb R-\{0\}$

$g(x)=\frac4{x+4}\implies\text{Dom}\,g=x\in\mathbb R-\{-4\}$

**JonathanEyoon** · August 30th 2007, 12:04 PM

Originally Posted by Krizalid

$f(x)=\frac2x\implies\text{Dom}\,f=x\in\mathbb R-\{0\}$

$g(x)=\frac4{x+4}\implies\text{Dom}\,g=x\in\mathbb R-\{-4\}$

So in composition, the domain is the original functions before they are put together?

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