What is the domain of f(3)=sqrt(x+1)+4 ?

The domain of f(3)=sqrt(x+1)+4 is x>=-1

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domain of f(3)=sqrt(x+1)+4

domain

f (3) = x + 1 + 4

x \geq - 1

Interval Notation

Solution steps

Find non-negative values for radicals:

x \geq - 1

The function domain

x \geq - 1

f (x) = \frac{∣ x ∣}{x ^{2} - 9}

f (x) = 8.9 x + 8.3

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

f (x) = x y^{2} + x y - 6 x - 3 = 0

f (x) = \frac{( x - 8 )}{( x + 2 )}

What is the domain of f(3)=sqrt(x+1)+4 ?
The domain of f(3)=sqrt(x+1)+4 is x>=-1