What is the domain of f(x)=sqrt((2x^2+8)/(x^2+9)) ?

The domain of f(x)=sqrt((2x^2+8)/(x^2+9)) is -infinity <x<infinity

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domain of f(x)=sqrt((2x^2+8)/(x^2+9))

domain

f (x) = \frac{2 x ^{2} + 8}{x ^{2} + 9}

- \infty < x < \infty

Interval Notation

(- \infty, \infty)

Solution steps

Find non-negative values for radicals:

- \infty < x < \infty

The function has no undefined points nor domain constraints. Therefore, the domain is

- \infty < x < \infty

f (x) = \frac{3 - ∣ x ∣}{x ^{2} - 9}

- x^{2} + 10 x

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

f (x) = \frac{x}{∣ x ∣}

lo g_{3} (4 - x^{2})

What is the domain of f(x)=sqrt((2x^2+8)/(x^2+9)) ?
The domain of f(x)=sqrt((2x^2+8)/(x^2+9)) is -infinity <x<infinity