What is the range of the function f(x) = √(x^2-x)/(x-|x|)?

[anemone]

Find the range of the function of $f$, where $f(x)=\dfrac{\sqrt{x^2-x}}{x-|x|}$.

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[anemone]

Re: Problem of the week #105 -March 31st, 2014

Congratulations to the following members for their correct solutions:

1. MarkFL
2. lfdahl

Solution from MarkFL:

Spoiler

We see that the domain of $f$ is $(-\infty,0)$.

We also find:

$$f'(x)=\frac{1}{2\sqrt{x^2-x}(x-|x|)}<0$$ for all $x$ in the domain, hence:

The upper bound of $f$ is:

$$\lim_{x\to-\infty}f(x)=-\lim_{x\to\infty}\frac{\sqrt{1+\frac{1}{x^2}}}{1+1}=-\frac{1}{2}$$

The lower bound of $f$ is:

$$\lim_{x\to0^{-}}f(x)=-\lim_{x\to0^{-}}\sqrt{\frac{1}{4}+\frac{1}{4x}}=-\infty$$

Thus, the range of $f$ is $$\left(-\infty,-\frac{1}{2} \right)$$

Solution from lfdahl:

Spoiler

The function: $f(x)=\frac{\sqrt{x^2-x}}{x-\left | x \right |}$ has the domain: $\mathbb{R_{-}} =\: \: \: \: ]-\infty ;0[$ because of the denominator being zero for $x \geq 0$. (Furthermore: $\sqrt{x^2-x}$ is only valid for $x\geq1$)
$f$ is strictly monotonic decreasing in its domain, because:

\[f'(x)= \frac{\mathrm{d} }{\mathrm{d} x}\frac{\sqrt{x^2-x}}{2x}=\frac{x}{4x^2\sqrt{x^2-x}} <0\; \; for \;\;x<0\]

I will check what happens when $x$ is close to zero using L’Hospital´s rule:
\[\lim_{x\rightarrow 0^-}f(x)=\lim_{x\rightarrow 0^-}\frac{\sqrt{x^2-x}}{x-\left | x \right |}= \lim_{x\rightarrow 0^-}\frac{\sqrt{x^2-x}}{2x}=\lim_{x\rightarrow 0^-}\frac{2x-1}{4\sqrt{x^2-x}}= -\infty\]
I need also to check what happens when $x \rightarrow -\infty$:
\[\lim_{x\rightarrow -\infty }f(x)=\lim_{x\rightarrow -\infty }\frac{\sqrt{x^2-x}}{x-\left | x \right |}= \lim_{x\rightarrow -\infty }-\frac{\sqrt{x^2-x}}{2\left | x \right |} \\\\ =\lim_{x\rightarrow -\infty }-\frac{\left | x \right |\sqrt{1-\frac{1}{x}}}{2\left | x \right |} = \lim_{x\rightarrow -\infty }-\frac{1}{2}\sqrt{1-\frac{1}{x}}=-\frac{1}{2}\]

Thus the range of $f$ is $]-\infty;-\frac{1}{2}[$