What is the domain of sqrt(4x^2-9) ?

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

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domain of sqrt(4x^2-9)

domain

4 x^{2} - 9

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

Interval Notation

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

Solution steps

Find non-negative values for radicals:

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

The function domain

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

f (x) = x^{2} - 6 x + 9

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

y = \frac{5 x}{2 x - 7}

\frac{x ^{2} - 3}{x ^{3} - 2 x ^{2} - x + 2}

f (x) = (x (x - 5))

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