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

The domain of f(x)=sqrt(16x^2-2) is x = 1/(2sqrt(2))

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

domain

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

x \leq - \frac{1}{2 2} or x \geq \frac{1}{2 2}

Interval Notation

(- \infty, - \frac{1}{2 2}] \cup [\frac{1}{2 2}, \infty)

Solution steps

Find non-negative values for radicals:

x \leq - \frac{1}{2 2} or x \geq \frac{1}{2 2}

The function domain

x \leq - \frac{1}{2 2} or x \geq \frac{1}{2 2}

f (x) = \frac{3}{- 3 - x}

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

2 x - 3 y + 4 x y - 7 = 0

f (x) = \frac{1}{2} x^{- \frac{2}{3}}

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

What is the domain of f(x)=sqrt(16x^2-2) ?
The domain of f(x)=sqrt(16x^2-2) is x<=-1/(2sqrt(2))\lor x>= 1/(2sqrt(2))