What is the domain of f(x)=sqrt(x^2-9),x>= 3 ?

The domain of f(x)=sqrt(x^2-9),x>= 3 is x>= 3

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domain of f(x)=sqrt(x^2-9),x>= 3

domain

f (x) = x^{2} - 9, x \geq 3

x \geq 3

Interval Notation

[3, \infty)

Solution steps

Set the domain to the given interval

x \geq 3

Find non-negative values for radicals:

x \leq - 3 or x \geq 3

Join real regions with the domain:

(x \leq - 3 or x \geq 3) and x \geq 3

x \geq 3

f (x) = 2 x + 2 x

f (x) = \frac{( 2 x - 5 )}{( x ^{2} - 3 x )}

\frac{4 x}{e ^{4 x}}

lo g_{3} (x + 9) + 6

What is the domain of f(x)=sqrt(x^2-9),x>= 3 ?
The domain of f(x)=sqrt(x^2-9),x>= 3 is x>= 3