What is the domain of f(x)=sqrt((2|x|-3)/(x^2+1)) ?

The domain of f(x)=sqrt((2|x|-3)/(x^2+1)) is x = 3/2

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domain of f(x)=sqrt((2|x|-3)/(x^2+1))

domain

f (x) = \frac{2 ∣ x ∣ - 3}{x ^{2} + 1}

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}

y = x - 11 + 5

f (x) = \frac{1 - 5 x}{2 + x}

5 x

f (x) = 6 x + 24

f (x) = arctan (2 x^{4})

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