What is the domain of f(x)=(\sqrt[3]{x}-1)/(2x^2+x-3) ?

The domain of f(x)=(\sqrt[3]{x}-1)/(2x^2+x-3) is x 1

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domain of f(x)=(\sqrt[3]{x}-1)/(2x^2+x-3)

domain

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

Interval Notation

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

Solution steps

Find undefined (singularity) points:

x = 1, x = - \frac{3}{2}

The function domain

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

y = - 8 - 4 x

f (x) = \frac{x + 10}{x}

y = 4 x^{3} - 30 x^{2} + 90 x + 100

F (x) = 5 x - 3

f (x) = sin (x^{2} + 2)

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