What is the domain of y=sqrt(4x^2-9)+1/(sqrt(x+2)) ?

The domain of y=sqrt(4x^2-9)+1/(sqrt(x+2)) is -2 = 3/2

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domain of y=sqrt(4x^2-9)+1/(sqrt(x+2))

domain

y = 4 x^{2} - 9 + \frac{1}{x + 2}

- 2 < x \leq - \frac{3}{2} or x \geq \frac{3}{2}

Interval Notation

(- 2, - \frac{3}{2}] \cup [\frac{3}{2}, \infty)

Solution steps

Find non-negative values for radicals:

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

Find undefined (singularity) points:

x = - 2

Combine real regions and undefined points for final function domain

- 2 < x \leq - \frac{3}{2} or x \geq \frac{3}{2}

What is the domain of y=sqrt(4x^2-9)+1/(sqrt(x+2)) ?
The domain of y=sqrt(4x^2-9)+1/(sqrt(x+2)) is -2<x<=-3/2 \lor x>= 3/2