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

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

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

domain

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

(- 5, 4), (0, - 2)

\frac{3 x ^{2}}{x - 5}

f (x) = 8 - 6 e^{x}

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

f (x) = - 8 x - 10

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