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Could somebody please show me how to write it as a piece-wise function?

$$g(x)=\frac{|x^3+x-2|}{|x|}$$

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closed as off-topic by mrtaurho, Lee David Chung Lin, ancientmathematician, GNUSupporter 8964民主女神 地下教會, Thomas Shelby Feb 12 at 17:00

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  • $\begingroup$ After your edit, you are asking one thing in the title and a different thing in the question! $\endgroup$ – TonyK Feb 5 at 11:38
  • $\begingroup$ Hi everybody! I'm so sorry for the confusion, I felt like my original question would be too much of a burden so I deleted it, and forgot about the title! Thanks for reminding me, my bad! $\endgroup$ – kim Feb 5 at 11:50
  • $\begingroup$ A hybrid function, is also a piece-wise function, if that helps. $\endgroup$ – kim Feb 5 at 11:51
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Note that $x^3+x-2=(x-1)(x^2+x+2)$ and $x^2+x+2$ is positive for all values of $x$, so the sign of $x^3+x-2$ is the same as the sign of $x-1$.

$g(x)$ is undefined at $x=0$ and $g(x)=0$ at $x=1$.Divide the rest of the $x$ axis into three regions:

  1. $x \lt 0$ where $x^3+x-2 \lt 0$ and $x \lt 0$
  2. $0 \lt x \lt 1$ where $x^3+x-2 \lt 0$ and $x \gt 0$
  3. $x \gt 1$ where $x^3+x-2 \gt 0$ and $x \gt 0$
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  • $\begingroup$ Hi gandalf61, could you please elaborate? I feel like I'm half way there... $\endgroup$ – kim Feb 5 at 11:55
  • $\begingroup$ Once you know the signs of the numerator and denominator in a particular region then you can re-write the expression for $g(x)$ in that region to remove the absolute value function. $\endgroup$ – gandalf61 Feb 5 at 13:34

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