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Find the domain and range of the function $$f(x)=\frac {x^2}{1-x}$$


My Attempt: It's obvious to see, from $f(x)$ that $x\neq 1$ or else the denominator equals $0$. So the domain is $$(-\infty,1)\cup (1,+\infty)$$

But for the range, I'm not quite sure. Solving for $y$ in terms of $x$ gives: $$y\cdot (1-x)=x^2\\y-xy-x^2=0\\x^2+xy-y=0$$ After this, I'm not sure what to do. This polynomial isn't factorable!

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  • $\begingroup$ Hint: the equation has real roots when $y^2+4y\ge0$. $\endgroup$ – Yves Daoust Oct 10 '16 at 13:40
  • $\begingroup$ @YvesDaoust Hm... But how is that going to help the range? I just need the possible $y$-values! $\endgroup$ – Frank Oct 10 '16 at 13:41
  • $\begingroup$ You should know, as you initiated this path yourself. $\endgroup$ – Yves Daoust Oct 10 '16 at 13:41
  • $\begingroup$ @YvesDaoust Oh wait, I see! The Discriminant is the only part that matters, so therefore, just set the Discriminant larger than of equal to $0$! $\endgroup$ – Frank Oct 10 '16 at 13:41
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You can find the range by studying the function, which we rewrite $f(x)=-(x+1)+1/(1-x)$ for convenience.

Then $f'(x)=-1+1/(1-x)^2$, which cancels at $x=0$, a minimum such that $f(0)=0$ and $x=2$, a maximum such that $f(2)=-4$. Hence the domain $(-\infty,-4]\cup[0,\infty)$.

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