What is the domain of f(x)=sqrt(3+(x^2-10)/(x-2)) ?

The domain of f(x)=sqrt(3+(x^2-10)/(x-2)) is (-3-sqrt(73))/2 = (sqrt(73)-3)/2

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domain of f(x)=sqrt(3+(x^2-10)/(x-2))

domain

f (x) = 3 + \frac{x ^{2} - 10}{x - 2}

\frac{- 3 - 73}{2} \leq x < 2 or x \geq \frac{73 - 3}{2}

Interval Notation

[\frac{- 3 - 73}{2}, 2) \cup [\frac{73 - 3}{2}, \infty)

Solution steps

Find non-negative values for radicals:

\frac{- 3 - 73}{2} \leq x < 2 or x \geq \frac{73 - 3}{2}

Find undefined (singularity) points:

x = 2

Combine real regions and undefined points for final function domain

\frac{- 3 - 73}{2} \leq x < 2 or x \geq \frac{73 - 3}{2}

y = x^{2} + 2 x + 18

f (x) = x^{2} - 6 x + 8

\frac{2 x}{x ^{2} + 2 x}

(4, 0), (20, 18)

(3 - x) e^{- x}

What is the domain of f(x)=sqrt(3+(x^2-10)/(x-2)) ?
The domain of f(x)=sqrt(3+(x^2-10)/(x-2)) is (-3-sqrt(73))/2 <= x<2\lor x>= (sqrt(73)-3)/2