What is the domain of 1/(sqrt(3+2x)) ?

The domain of 1/(sqrt(3+2x)) is x>-3/2

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domain of 1/(sqrt(3+2x))

domain

\frac{1}{3 + 2 x}

x > - \frac{3}{2}

Interval Notation

(- \frac{3}{2}, \infty)

Solution steps

Find non-negative values for radicals:

x \geq - \frac{3}{2}

Find undefined (singularity) points:

x = - \frac{3}{2}

Combine real regions and undefined points for final function domain

x > - \frac{3}{2}

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

(1664 - 64 y^{2}) y^{2}

f (x) = \frac{x - 1}{x + 5}

x^{\frac{7}{2}} - 4 x^{2}

f (x) = \frac{8 + x ^{4}}{x ^{2} - x ^{4}}