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

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

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

domain

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

x < 1 or x > 1

Interval Notation

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

Solution steps

Find undefined (singularity) points:

x = 1

The function domain

x < 1 or x > 1

f (x) = 3 x^{2} + 1

\frac{- 2}{x - 2}

f (x) = - 3 + ln (x)

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

\frac{∣ x ∣}{∣ x ∣ - 5}

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