What is the domain of f(x)=sqrt(-8+1/2 x) ?

The domain of f(x)=sqrt(-8+1/2 x) is x>= 16

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domain of f(x)=sqrt(-8+1/2 x)

domain

f (x) = - 8 + \frac{1}{2} x

x \geq 16

Interval Notation

[16, \infty)

Solution steps

Find non-negative values for radicals:

x \geq 16

The function domain

x \geq 16

g (x) = (\frac{1}{2})^{x}

f (x) = \frac{25 - 5 x}{4}

f (x) = 3 x^{2} - 2 x + 5

f (x) = x^{2} y^{2} - 4 x^{2} - 4 y^{2} = 0

f (x) = \frac{x - 5}{2 x + 1}

What is the domain of f(x)=sqrt(-8+1/2 x) ?
The domain of f(x)=sqrt(-8+1/2 x) is x>= 16