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

The domain of f(x)= 1/(x-2)+sqrt(3x) is 0 2

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

domain

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

0 \leq x < 2 or x > 2

Interval Notation

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

Solution steps

Find non-negative values for radicals:

x \geq 0

Find undefined (singularity) points:

x = 2

Combine real regions and undefined points for final function domain

0 \leq x < 2 or x > 2

f (x) = - lo g_{\frac{1}{5}} (x + 6) - 5

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

f (x) = - 0. 6^{(x + 2)} + 3

\frac{x - 2}{x + 6}

f (x) = \frac{x}{3 x + 9}

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