What is the domain of sqrt(2x^2-17) ?

The domain of sqrt(2x^2-17) is x = sqrt(17/2)

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domain of sqrt(2x^2-17)

domain

2 x^{2} - 17

x \leq - \frac{17}{2} or x \geq \frac{17}{2}

Interval Notation

(- \infty, - \frac{17}{2}] \cup [\frac{17}{2}, \infty)

Solution steps

Find non-negative values for radicals:

x \leq - \frac{17}{2} or x \geq \frac{17}{2}

The function domain

x \leq - \frac{17}{2} or x \geq \frac{17}{2}

f (x) = 2 (2 x + 8) + 8

f (x) = - 6 - 2 x

y = - 4 x - 9, (0, 3)

y = - 5 x - 1

x^{3} - 6

What is the domain of sqrt(2x^2-17) ?
The domain of sqrt(2x^2-17) is x<=-sqrt(17/2)\lor x>= sqrt(17/2)