How do you find the domain of $y=4/(2(x-2)^2-1 )$ ?

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Answer 1 · 2017-03-18T13:30:55

The domain is $D_y=RR-{2-1/sqrt2,2+1/sqrt2}$

$y=4/(2(x-2)^2-1)$

As we cannot divide by $0$ ,

$2(x-2)^2-1!=0$

$2(x-2)^2!=1$

$(x-2)^2!=1/2$

$(x-2)!=+-1/sqrt2$

$x!=2+-1/sqrt2$

The domain of $y$ is $D_y=RR-{2-1/sqrt2,2+1/sqrt2}$

Answer 2 · 2017-03-18T13:33:58

$x in RR - {2-1/sqrt(2), 2+1/sqrt(2)}$

$y=4/(2(x-2)^2-1)$ is defined for all values of $x$ for which

$color(white)("XXX")2(x-2)^2-1 !=0$

That is for all values except
$color(white)("XXX")(x-2)^2=1/2$

$color(white)("XXX")x-2=+-1/sqrt(2)$

$color(white)("XXX")x=2+-1/sqrt(2)$

How do you find the domain of y=4/(2(x-2)^2-1 )y=4/(2(x-2)^2-1 )?