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

The domain of f(x)=sqrt((16-x^2)(x+2)) is x<=-4\lor-2<= x<= 4

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

domain

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

x \leq - 4 or - 2 \leq x \leq 4

Interval Notation

(- \infty, - 4] \cup [- 2, 4]

Solution steps

Find non-negative values for radicals:

x \leq - 4 or - 2 \leq x \leq 4

The function domain

x \leq - 4 or - 2 \leq x \leq 4

f (x) = \frac{x ^{2} - 2 x - 15}{x + 3}

f (x) = \frac{4 x}{x ^{2} + 1}

f (x) = \frac{x ^{3} - 27}{x ^{2} - 8 x + 15}

e^{x + 2}

x^{2} + 4

What is the domain of f(x)=sqrt((16-x^2)(x+2)) ?
The domain of f(x)=sqrt((16-x^2)(x+2)) is x<=-4\lor-2<= x<= 4