What is the domain of f(x)=sqrt(2x^2-x-6+2) ?

The domain of f(x)=sqrt(2x^2-x-6+2) is x = (sqrt(33)+1)/4

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domain of f(x)=sqrt(2x^2-x-6+2)

domain

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

x \leq \frac{- 33 + 1}{4} or x \geq \frac{33 + 1}{4}

Interval Notation

(- \infty, \frac{- 33 + 1}{4}] \cup [\frac{33 + 1}{4}, \infty)

Solution steps

Find non-negative values for radicals:

x \leq \frac{- 33 + 1}{4} or x \geq \frac{33 + 1}{4}

The function domain

x \leq \frac{- 33 + 1}{4} or x \geq \frac{33 + 1}{4}

What is the domain of f(x)=sqrt(2x^2-x-6+2) ?
The domain of f(x)=sqrt(2x^2-x-6+2) is x<= (-sqrt(33)+1)/4 \lor x>= (sqrt(33)+1)/4