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Question Number 83653 by jagoll last updated on 05/Mar/20

find range of function   y= (4/((x^2 −4)))

$$\mathrm{find}\:\mathrm{range}\:\mathrm{of}\:\mathrm{function}\: \\ $$$$\mathrm{y}=\:\frac{\mathrm{4}}{\left({x}^{\mathrm{2}} −\mathrm{4}\right)} \\ $$

Answered by MJS last updated on 05/Mar/20

x=±2(√((y+1)/y))  ⇒ y≤−1∨y>0

$${x}=\pm\mathrm{2}\sqrt{\frac{{y}+\mathrm{1}}{{y}}} \\ $$$$\Rightarrow\:{y}\leqslant−\mathrm{1}\vee{y}>\mathrm{0} \\ $$

Commented by jagoll last updated on 05/Mar/20

yess

$$\mathrm{yess} \\ $$

Answered by jagoll last updated on 05/Mar/20

domain x ≠ ± 2  ⇔ x^2 −4 = (4/y) ⇒ x= ± (√(4+(4/y)))  x = ± (√((4y+4)/y)) , ((4y+4)/y)≥0  y ≤ −1 ∨ y > 0

$$\mathrm{domain}\:{x}\:\neq\:\pm\:\mathrm{2} \\ $$$$\Leftrightarrow\:{x}^{\mathrm{2}} −\mathrm{4}\:=\:\frac{\mathrm{4}}{{y}}\:\Rightarrow\:{x}=\:\pm\:\sqrt{\mathrm{4}+\frac{\mathrm{4}}{{y}}} \\ $$$${x}\:=\:\pm\:\sqrt{\frac{\mathrm{4y}+\mathrm{4}}{\mathrm{y}}}\:,\:\frac{\mathrm{4y}+\mathrm{4}}{\mathrm{y}}\geqslant\mathrm{0} \\ $$$$\mathrm{y}\:\leqslant\:−\mathrm{1}\:\vee\:\mathrm{y}\:>\:\mathrm{0} \\ $$

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