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C3 functions

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B_9710
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Original post by thegreatwhale
lol still don't know where the 3 comes from and where the 8x3−18x^3 -1 went


8x3−11−2x3≡−4(1−2x3)+31−2x3 \displaystyle \frac{8x^3-1}{1-2x^3} \equiv \frac{-4(1-2x^3)+3}{1-2x^3} .
You can separate this into 2 separate fractions −4(1−2x3)1−2x3+31−2x3=−4+31−2x3 \displaystyle \frac{-4(1-2x^3)}{1-2x^3}+\frac{3}{1-2x^3} = -4+\frac{3}{1-2x^3} .
Original post by thegreatwhale
woah where did that come from?? o..o
i understand 11−2x3=(1−2x3)−1\dfrac{1}{1-2x^3} =(1-2x^3)^{-1}

but i'm a little unsure of where the 3 comes from ad i'm assuming the -4 is to balance things out


8x3−11−2x3=−4(1−2x3)+31−2x3=−4×1−2x31−2x3+31−2x3=3(1−2x3)−1−4\dfrac{8x^3-1}{1-2x^3}=\dfrac{-4(1-2x^3)+3}{1-2x^3}=-4 \times \dfrac{1-2x^3}{1-2x^3}+\dfrac{3}{1-2x^3}=3(1-2x^3)^{-1}-4

It's not necessary. I noticed there there was an x^3 in the numerator and denominator and no other powers of x so I simplified and differentiated at the time. I wouldn't have used polynomial long division for a calculus problem if it were more complicated.
(edited 7 years ago)
Original post by fefssdf
I'm actually so confused right now

me 2
Original post by B_9710
8x3−11−2x3≡−4(1−2x3)+31−2x3 \displaystyle \frac{8x^3-1}{1-2x^3} \equiv \frac{-4(1-2x^3)+3}{1-2x^3} .
You can separate this into 2 separate fractions −4(1−2x3)1−2x3+31−2x3=−4+31−2x3 \displaystyle \frac{-4(1-2x^3)}{1-2x^3}+\frac{3}{1-2x^3} = -4+\frac{3}{1-2x^3} .


Original post by Kvothe the arcane
8x3−11−2x3=−4(1−2x3)+31−2x3=−4×1−2x31−2x3+31−2x3=3(1−2x3)−1−4\dfrac{8x^3-1}{1-2x^3}=\dfrac{-4(1-2x^3)+3}{1-2x^3}=-4 \times \dfrac{1-2x^3}{1-2x^3}+\dfrac{3}{1-2x^3}=3(1-2x^3)^{-1}-4


ah awesome ^-^ for got you could "take out a common factor from anything as long as u balance it" thanks :biggrin:

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