anonymous
  • anonymous
f(x) = 5x + 3; g(x) = 6x - 5 Find f/g. 5x + 3/6x - 5 domain {x|x ≠3/5 or 5x + 3/6x - 5 domain {x|x ≠ 5/6 ?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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Luigi0210
  • Luigi0210
Do you know how to find the domain?
anonymous
  • anonymous
I'm pretty sure it's the second one because the bottom can't equal to 0 right?
Luigi0210
  • Luigi0210
You are correct sir!

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More answers

anonymous
  • anonymous
Thank you! I just don't want to take chances with picking the wrong one XD Could you maybe help me with one more question?
Luigi0210
  • Luigi0210
Sure.
anonymous
  • anonymous
Find f(x) and g(x) so the function can be expressed as y = f(g(x)). y = 8/x^2 + 4 I want to say g(x)= 1/x f(x)= 8/x +4 but it doesn't look right.
Luigi0210
  • Luigi0210
Okay, there are many possibilities for something like this. I would say the easiest way, since it's a fraction, would be to make the denominator an x. and make g(x) whatever is in the denominator
Luigi0210
  • Luigi0210
If you plug in what you got we would get this: \[\frac{8}{\frac{1}{x}+4}\]
Luigi0210
  • Luigi0210
Are you with me so far or did I lose you? :/
anonymous
  • anonymous
I'm a bit lost XD But does this look alright? I think I got it on my own. g(x)= 1/x^3 f(x)= 8x +4x^3
Luigi0210
  • Luigi0210
Hm, well plug it in, where ever you see an x for f(x) plug in g(x)
anonymous
  • anonymous
|dw:1377715598143:dw| Because the x's cancel each other out right?
Luigi0210
  • Luigi0210
No, you don't keep the x, you plug it INTO the x.
Luigi0210
  • Luigi0210
If you had this: f(3)=x^2-1 What would you do?
anonymous
  • anonymous
Oh. Then carry on XD
anonymous
  • anonymous
3^2-1
Luigi0210
  • Luigi0210
Exactly, and f(g(x)) is the same concept, expect you're using equations instead of numbers.
anonymous
  • anonymous
Would it be this simple then? g(x)= x^2 f(x)= 8/x +4
Luigi0210
  • Luigi0210
|dw:1377716036694:dw| The simplest way to think of it, like I said earlier was to just separate them into two different equations
Luigi0210
  • Luigi0210
And actually the way you have it is right too :)
Luigi0210
  • Luigi0210
So yes it's that simple :P
anonymous
  • anonymous
Thanks so much XD
Luigi0210
  • Luigi0210
Anytime :)

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