What is the domain of f(x)=\sqrt[4]{3/(9x^2-6x-1)} ?

The domain of f(x)=\sqrt[4]{3/(9x^2-6x-1)} is x (1+sqrt(2))/3

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domain of f(x)=\sqrt[4]{3/(9x^2-6x-1)}

domain

x < \frac{- 2 + 1}{3} or x > \frac{1 + 2}{3}

Interval Notation

(- \infty, \frac{- 2 + 1}{3}) \cup (\frac{1 + 2}{3}, \infty)

Solution steps

Find non-negative values for radicals:

x < \frac{- 2 + 1}{3} or x > \frac{2 + 1}{3}

Find undefined (singularity) points:

x = \frac{1 + 2}{3}, x = \frac{1 - 2}{3}

Combine real regions and undefined points for final function domain

x < \frac{- 2 + 1}{3} or x > \frac{1 + 2}{3}

f (x) = (x - 3)^{2} + 2

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

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

y = \frac{1}{ln ( x )}

f (x) = x + 64

What is the domain of f(x)=\sqrt[4]{3/(9x^2-6x-1)} ?
The domain of f(x)=\sqrt[4]{3/(9x^2-6x-1)} is x<(-sqrt(2)+1)/3 \lor x>(1+sqrt(2))/3