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

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

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

domain

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

x < 1 or x > 1

Interval Notation

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

Solution steps

Find non-negative values for radicals:

- \infty < x < \infty

Find undefined (singularity) points:

x = 1

Combine real regions and undefined points for final function domain

x < 1 or x > 1

\frac{2 x}{10 x + 1}

y = (\frac{4}{5})^{x}

\frac{x + 6}{x - 6}

f (x) = - e^{\frac{1}{5} x}

f (x) = \frac{x ^{2} + 4 x}{5 x ^{2} + 36 x + 64}

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