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

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

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

domain

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

x < - 1 or x \geq \frac{3}{2}

Interval Notation

(- \infty, - 1) \cup [\frac{3}{2}, \infty)

Solution steps

Find non-negative values for radicals:

x < - 1 or x \geq \frac{3}{2}

Find undefined (singularity) points:

x = - 1

Combine real regions and undefined points for final function domain

x < - 1 or x \geq \frac{3}{2}

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

f (x) = \frac{x - 5}{x ^{2} - 6 x + 8}

f (x) = \frac{6}{x + 3} \cdot \frac{7}{x}

f (x) = ∣ x ∣^{2} - 16

f (x) = x^{\frac{4}{5}} (x^{2} - 9)

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