What is the domain of [(x/(x^2-1))(sqrt(1-2x))]^2 ?

The domain of [(x/(x^2-1))(sqrt(1-2x))]^2 is x<-1\lor-1<x<= 1/2

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domain of [(x/(x^2-1))(sqrt(1-2x))]^2

domain

[(\frac{x}{x ^{2} - 1}) (1 - 2 x)]^{2}

x < - 1 or - 1 < x \leq \frac{1}{2}

Interval Notation

(- \infty, - 1) \cup (- 1, \frac{1}{2}]

Solution steps

Find non-negative values for radicals:

x \leq \frac{1}{2}

Find undefined (singularity) points:

x = 1, x = - 1

Combine real regions and undefined points for final function domain

x < - 1 or - 1 < x \leq \frac{1}{2}

f (x) = x^{7} + x^{6} - x^{2}

f (x) = - x + 9

y = 2^{x - 2}

\frac{2}{3} - \frac{2}{3 ( 3 x + 1 )}

f (x) = 5 x + 20

What is the domain of [(x/(x^2-1))(sqrt(1-2x))]^2 ?
The domain of [(x/(x^2-1))(sqrt(1-2x))]^2 is x<-1\lor-1<x<= 1/2