What is the domain of f(x)=sqrt(4x+52) ?

The domain of f(x)=sqrt(4x+52) is x>=-13

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domain of f(x)=sqrt(4x+52)

domain

f (x) = 4 x + 52

x \geq - 13

Interval Notation

[- 13, \infty)

Solution steps

Find non-negative values for radicals:

x \geq - 13

The function domain

x \geq - 13

\frac{1}{x} + 10

\frac{x ^{2} - 4}{2 x ^{2}}

f (x) = \frac{6}{x + 4} - \frac{1}{2 - x}

f (x) = - ln (x + 1)

(0, - 2), (5, 3)

What is the domain of f(x)=sqrt(4x+52) ?
The domain of f(x)=sqrt(4x+52) is x>=-13