What is the domain of (x^3+3x^2+1)/(4x(x-1)^2) ?

The domain of (x^3+3x^2+1)/(4x(x-1)^2) is x 1

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domain of (x^3+3x^2+1)/(4x(x-1)^2)

domain

\frac{x ^{3} + 3 x ^{2} + 1}{4 x ( x - 1 ) ^{2}}

x < 0 or 0 < x < 1 or x > 1

Interval Notation

(- \infty, 0) \cup (0, 1) \cup (1, \infty)

Solution steps

Find undefined (singularity) points:

x = 0, x = 1

The function domain

x < 0 or 0 < x < 1 or x > 1

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

f (x) = \frac{4 x ^{2} - 36}{x ^{2} - 2 x - 8}

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

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

f (x) = \frac{x ^{2}}{3}

What is the domain of (x^3+3x^2+1)/(4x(x-1)^2) ?
The domain of (x^3+3x^2+1)/(4x(x-1)^2) is x<0\lor 0<x<1\lor x>1