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

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

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

domain

f (x) = \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{1}{( y - 2 ) + ( y - 8 )}

f (x) = - \frac{3}{4} x + 2

f (x) = ∣ x ∣ + 4

\frac{x - 2}{5 - x}

f (x) = \frac{( x + ∣ x ∣ )}{x}

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