What is the domain of y=((x+7)^2)/(sqrt(6x-2)) ?

The domain of y=((x+7)^2)/(sqrt(6x-2)) is x> 1/3

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domain of y=((x+7)^2)/(sqrt(6x-2))

domain

y = \frac{( x + 7 ) ^{2}}{6 x - 2}

x > \frac{1}{3}

Interval Notation

(\frac{1}{3}, \infty)

Solution steps

Find non-negative values for radicals:

x \geq \frac{1}{3}

Find undefined (singularity) points:

x = \frac{1}{3}

Combine real regions and undefined points for final function domain

x > \frac{1}{3}

\frac{( x - 1 ) ( x + 9 )}{x ^{2} - 9}

f (x) = ln ∣ x - 6 ∣

f (x) = ln (1 + 2 x^{2})

\frac{7 x}{3 x - 1}

f (x) = \frac{2 x - 4}{3 x - 5}

What is the domain of y=((x+7)^2)/(sqrt(6x-2)) ?
The domain of y=((x+7)^2)/(sqrt(6x-2)) is x> 1/3