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Have this function:

$$f(x) = \frac{x}{\sqrt{x^2+2x}}$$

And for whatever reason I decide to divide both the numerator and denominator by $x$:

$$f(x) = \frac{1}{\sqrt{1+\frac{2}{x}}}$$

My understanding is that the function is equivalent because dividing both parts by the exact same value doesn't change it algebraically.

And yet, when I look at the graphs, something strange happens with minus infinity:

The original:

enter image description here

The modified one:

enter image description here

The positive domain is the same it seems - it only appears flipped in the negative domain. My original understanding was that if the functions were algebraically the same, the graphs should be identical. But it appears that my division messed it up somehow.

Why is the negative domain flipped because of my division?

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Because, for all $x\in(-\infty, -2)\cup (0,\infty)$, $\sqrt{x^2+2x}=\lvert x\rvert\sqrt{1+\frac2x}$, as opposed to your suggestion $x\sqrt{1+\frac2x}$.

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  • $\begingroup$ Which means that if instead of dividing the numerator and denominator by $x$ I divide them by $|x|$, it will work as expected? (I just tested and yeah the graphs now look the same, just wanted to confirm if this was a valid way to do it) $\endgroup$ – Zol Tun Kul Sep 11 at 5:49
  • $\begingroup$ Yes.${}{}{}{}{}$ $\endgroup$ – Gae. S. Sep 11 at 6:07
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I think you have done the following steps:

$ \frac{\sqrt{x^2+2x}}{x}= \sqrt{\frac{x^2+2x}{x^2}}.$

But this is only valid for $x>0$, since $ \sqrt{x^2}=|x|.$

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You are not dividing both the numerator and denominator by $x$, but by $|x|$(that is, $\sqrt{x^2}$).

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By forcibly putting $\frac1x$ into the square root, you need to consider $\pm\sqrt{\frac{1}{x^2}}$.

Otherwise, the below is correct: $$f(x)=\frac{x}{\sqrt{x^2+2x}}=\frac{1}{\frac1x\sqrt{x^2+2x}}.$$

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