What is the domain of f(x)=sqrt(1/(x^2)+1/(2x-1)) ?

The domain of f(x)=sqrt(1/(x^2)+1/(2x-1)) is -1-sqrt(2) 1/2

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domain of f(x)=sqrt(1/(x^2)+1/(2x-1))

domain

f (x) = \frac{1}{x ^{2}} + \frac{1}{2 x - 1}

- 1 - 2 \leq x < 0 or 0 < x \leq - 1 + 2 or x > \frac{1}{2}

Interval Notation

[- 1 - 2, 0) \cup (0, - 1 + 2] \cup (\frac{1}{2}, \infty)

Solution steps

Find non-negative values for radicals:

- 1 - 2 \leq x < 0 or 0 < x \leq - 1 + 2 or x > \frac{1}{2}

Find undefined (singularity) points:

x = 0, x = \frac{1}{2}

Combine real regions and undefined points for final function domain

- 1 - 2 \leq x < 0 or 0 < x \leq - 1 + 2 or x > \frac{1}{2}

f (x) = \frac{2 y}{9 + y ^{2}}

f (x) = x + 4 - 8

f (x) = t + 7

f (x) = - x - 6

f (x) = \frac{x + 3}{x - 5}

What is the domain of f(x)=sqrt(1/(x^2)+1/(2x-1)) ?
The domain of f(x)=sqrt(1/(x^2)+1/(2x-1)) is -1-sqrt(2)<= x<0\lor 0<x<=-1+sqrt(2)\lor x> 1/2