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September 2nd, 2014, 09:45 AM   #1
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range of y

Find range of
y=(1-x^2)/(1+x^2)

Last edited by skipjack; September 2nd, 2014 at 11:47 AM.
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September 2nd, 2014, 10:16 AM   #2
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$$\lim_{x \to -\infty} y = \lim_{x \to +\infty} y = -1 \\ \frac{\mathbb{d}y}{\mathbb{d}x} = 0 \implies x=0 \\
y(0) = 1$$
and $y$ is bounded everywhere.

So the range of $y$ is $(-1, 1]$.
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September 2nd, 2014, 11:59 AM   #3
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As y = -1 + 2/(1 + x²) = 1 - 2x²/(1 + x²), y ∈ (-1,1].
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