What is the domain of f(x)=sqrt(x^2-4)+1/(x^2-9)+x ?

The domain of f(x)=sqrt(x^2-4)+1/(x^2-9)+x is x 3

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domain of f(x)=sqrt(x^2-4)+1/(x^2-9)+x

domain

f (x) = x^{2} - 4 + \frac{1}{x ^{2} - 9} + x

x < - 3 or - 3 < x \leq - 2 or 2 \leq x < 3 or x > 3

Interval Notation

(- \infty, - 3) \cup (- 3, - 2] \cup [2, 3) \cup (3, \infty)

Solution steps

Find non-negative values for radicals:

x \leq - 2 or x \geq 2

Find undefined (singularity) points:

x = 3, x = - 3

Combine real regions and undefined points for final function domain

x < - 3 or - 3 < x \leq - 2 or 2 \leq x < 3 or x > 3

0.05 x + 20 + \frac{125}{x}

f (x) = 6 y - 1248

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

- \frac{11}{( 5 + t ) ^{2}}

f (x) = \frac{( 2 - x )}{( x + 1 ) ^{5}}

What is the domain of f(x)=sqrt(x^2-4)+1/(x^2-9)+x ?
The domain of f(x)=sqrt(x^2-4)+1/(x^2-9)+x is x<-3\lor-3<x<=-2\lor 2<= x<3\lor x>3