What is the domain of f(x)=sqrt(2x^2-4x-16) ?

The domain of f(x)=sqrt(2x^2-4x-16) is x = 4

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domain of f(x)=sqrt(2x^2-4x-16)

domain

f (x) = 2 x^{2} - 4 x - 16

x \leq - 2 or x \geq 4

Interval Notation

(- \infty, - 2] \cup [4, \infty)

Solution steps

Find non-negative values for radicals:

x \leq - 2 or x \geq 4

The function domain

x \leq - 2 or x \geq 4

f (x) = 6 + x^{2}

f (x) = ln (x^{2} - 1) + 2

f (x) = \frac{x + 3}{x ^{2} + 4 x - 5}

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

f (x) = - x^{2} - 4 x - 5

What is the domain of f(x)=sqrt(2x^2-4x-16) ?
The domain of f(x)=sqrt(2x^2-4x-16) is x<=-2\lor x>= 4