What is the domain of f(x)=sqrt(2x^2-9x-5) ?

The domain of f(x)=sqrt(2x^2-9x-5) is x = 5

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domain of f(x)=sqrt(2x^2-9x-5)

domain

f (x) = 2 x^{2} - 9 x - 5

x \leq - \frac{1}{2} or x \geq 5

Interval Notation

(- \infty, - \frac{1}{2}] \cup [5, \infty)

Solution steps

Find non-negative values for radicals:

x \leq - \frac{1}{2} or x \geq 5

The function domain

x \leq - \frac{1}{2} or x \geq 5

f (x) = - 4 x + 6 - 15

f (x) = lo g_{2} (x + 1) - 3

x - 2 + 4 - x

f (x) = \frac{x}{2 x ^{2} - 6 x + 4}

f (x) = (x + 1) 2

What is the domain of f(x)=sqrt(2x^2-9x-5) ?
The domain of f(x)=sqrt(2x^2-9x-5) is x<=-1/2 \lor x>= 5