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

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

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

domain

f (x) = \frac{1}{x} + \frac{1}{x - 3} + \frac{1}{x - 2}

x < 0 or 0 < x < 2 or 2 < x < 3 or x > 3

Interval Notation

(- \infty, 0) \cup (0, 2) \cup (2, 3) \cup (3, \infty)

Solution steps

Find undefined (singularity) points:

x = 0, x = 3, x = 2

The function domain

x < 0 or 0 < x < 2 or 2 < x < 3 or x > 3

y = x^{3} - 2 x^{2} - 4 x + 3

f (x) = 4 x - 12

f (x) = \frac{( x + 2 )}{( x ^{2} - x - 6 )}

\frac{3 x ^{2} + 1}{x ^{2} - 2 x - 3}

tan^{3} (x) d x

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