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Precalculus
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How would i find the range without a calculator or graph for (4x)/((x^2)+2)? I know what it is but i have to show the work and i dont know how to algebraically
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\[\frac{4x}{x^2+2}=y\] without using calc you can find the range by solving for \(x\)
what would i do after this? i got 4x=yx^2+2y then -2y=yx^2-4x how would i solve for x with more than one x?
\[ 4x=yx^2+2y\] \[yx^2-4x-2y=0\] use \(\frac{-b\pm\sqrt{b^2-4ac}}{2a}\)with \[a=y,b=-4,c=-2y\] and basically you have to make sure that \(b^2-4ac\geq 0\) otherwise it is notdefined
\[16-4\times y\times (-2y)\geq 0\] \[16+8y^2\geq 0\] etc
i made some mistake
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\[yx^2-4x+2y=0\] \[16-8y^2\geq 0\] etc
\[2-y^2\geq0\] \[-\sqrt{2}\leq y\leq \sqrt{2}\]
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