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September 5th, 2011, 02:26 PM   #1
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Precalculus

Hello Everyone. Among plenty more problems in my precalculus class that I am having trouble with, this is one that has me the most baffled. If someone could please solve it and tell me how they did so I would be extremely grateful.

Here it is:

f(x) = 2x-5x ; g(x) = 2x/x-5 (two x over x minus 5)

All I am asked is to find fg

I have been trying to solve this for a good while, and honestly, I am not sure if it is asking me to multiply the two functions, or to find f(g). Either way, I tried doing both and got no answer that was anywhere near the 5 choices given by my instructor, which are listed below. (note, 2x^5 means 2x to the 5th power etc.. If someone knows the ASCII code for "superscripts 4 and 5", please share them with me).

1. 2x^5+25x-75x
2. 2x^5-25x-75x
3. 2x^5+16x^4+25x-40x-75x
4. 2x^5+16x^4-35x-40x-75x
5. 2x^5+16x^4+25x+40x+75x

These are the answer choices given to me, and neither me or my friend/classmate have been able to make much sense of this. If anyone out there can help, that would be awesome. Thanks, and have a great day!
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September 5th, 2011, 02:49 PM   #2
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Re: Precalculus

Something is screwed up. When you multiply you get 4x^3 - 2x^2. f(g) leads to a rational function with both numerator and denominator has cubic polynomials.
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September 5th, 2011, 03:04 PM   #3
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Re: Precalculus

Wow..thank you for pointing that out. I've literally been doing math homework for 2 days straight, I think it's time for a break. But regardless, here is the CORRECT problem. I was trying to figure it out last night and couldn't.

f(x) = 2x-5x
g(x) = x+8x+15

These are correct, I double checked twice, as well as the answer choices I gave in my original post.
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September 5th, 2011, 04:32 PM   #4
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Re: Precalculus

It looks like f * g.

(2x - 5x)(x + 8x + 15) = 2x^5 + 16x^4 + 25x^3 - 40x^2 - 75x, choice number 3.
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