What is the domain of f(x)=sqrt(x^2+12x+32) ?

The domain of f(x)=sqrt(x^2+12x+32) is x =-4

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domain of f(x)=sqrt(x^2+12x+32)

domain

f (x) = x^{2} + 12 x + 32

x \leq - 8 or x \geq - 4

Interval Notation

(- \infty, - 8] \cup [- 4, \infty)

Solution steps

Find non-negative values for radicals:

x \leq - 8 or x \geq - 4

The function domain

x \leq - 8 or x \geq - 4

y = 4 x^{2} + 7 x - 19

ln (5 - t^{2})

f (x) = \frac{1}{x + 3} - 5

\frac{x + 8}{2}

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

What is the domain of f(x)=sqrt(x^2+12x+32) ?
The domain of f(x)=sqrt(x^2+12x+32) is x<=-8\lor x>=-4