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

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

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

domain

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

x < 0 or 0 < x < 2 or x > 2

Interval Notation

(- \infty, 0) \cup (0, 2) \cup (2, \infty)

Solution steps

Find undefined (singularity) points:

x = 0, x = 2

The function domain

x < 0 or 0 < x < 2 or x > 2

\frac{3 - x}{x ^{2} - 4}

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

x^{3} + (4 x^{2} + 17 x - 15)

f (x) = \frac{x + 5}{x - 6}

f (x) = x - 4 - 2

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