What is the domain of f(x)=(sqrt(x-3))+(x^2-16) ?

The domain of f(x)=(sqrt(x-3))+(x^2-16) is x>= 3

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domain of f(x)=(sqrt(x-3))+(x^2-16)

domain

f (x) = (x - 3) + (x^{2} - 16)

x \geq 3

Interval Notation

[3, \infty)

Solution steps

Find non-negative values for radicals:

x \geq 3

The function domain

x \geq 3

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

(x)^{2}

f (x) = - 3 (2)^{x + 4}

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

What is the domain of f(x)=(sqrt(x-3))+(x^2-16) ?
The domain of f(x)=(sqrt(x-3))+(x^2-16) is x>= 3