How do you find the domain and range of $y = 1/(x^2 - 2)$ ?

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Answer 1 · 2017-11-06T11:41:22

Domain: $x|oo,-sqrt(2))uu(-sqrt(2),sqrt(2))uu(sqrt(2),oo)$
Range: $y| (-oo,0)uu(0,oo)$

$y=1/(x^2-2) = 1/((x+sqrt2)(x-sqrt2))$ . Function is undefined

if denominator is zero.So function is undefined at $x=sqrt2$

and at $x=-sqrt2$ . Domain : Any real number of $x$ except

$x=+-sqrt2$ . Domain: $x in RR | x !=+-sqrt2$ . In interval notation

expressed as $x|(-oo , -sqrt(2))uu(-sqrt(2),sqrt(2))uu(sqrt(2),oo)$ .

Range: Any real number of $y$ except $y=0$

Range: $y in RR | y !=0$ .In interval notation expressed as

$y| (-oo,0)uu(0,oo)$

graph{1/(x^2-2) [-10, 10, -5, 5]} [Ans]

How do you find the domain and range of y = 1/(x^2 - 2)y = 1/(x^2 - 2)?