What is the domain of f(x)=sqrt(x+9) ?

The domain of f(x)=sqrt(x+9) is x>=-9

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domain of f(x)=sqrt(x+9)

domain

f (x) = x + 9

x \geq - 9

Interval Notation

[- 9, \infty)

Solution steps

Find non-negative values for radicals:

x \geq - 9

The function domain

x \geq - 9

y = 10 x^{2}

f (x) = cos (x - \frac{π}{2})

(2 - x^{2}) (x^{2} - 9)

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