What is the domain of h(x)=(x^2+5)/(x-1) ?

The domain of h(x)=(x^2+5)/(x-1) is x 1

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domain of h(x)=(x^2+5)/(x-1)

domain

h (x) = \frac{x ^{2} + 5}{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) = \frac{1}{∣ x ∣ - x}

f (x) = x^{n}

f (x) = x + 9 + 4

\frac{x ^{2} - 1}{x ^{2} - 16}

7 x + 2

What is the domain of h(x)=(x^2+5)/(x-1) ?
The domain of h(x)=(x^2+5)/(x-1) is x<1\lor x>1