What is the domain of f(x)=sqrt(12x^2-5x-150) ?

The domain of f(x)=sqrt(12x^2-5x-150) is x = 15/4

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domain of f(x)=sqrt(12x^2-5x-150)

domain

f (x) = 12 x^{2} - 5 x - 150

x \leq - \frac{10}{3} or x \geq \frac{15}{4}

Interval Notation

(- \infty, - \frac{10}{3}] \cup [\frac{15}{4}, \infty)

Solution steps

Find non-negative values for radicals:

x \leq - \frac{10}{3} or x \geq \frac{15}{4}

The function domain

x \leq - \frac{10}{3} or x \geq \frac{15}{4}

f (x) = ln (0.5) - 0.5

f (x) = 3 + tan (x + \frac{π}{2})

h (x) = 289 - x^{2}

\frac{x}{7} + \frac{5}{7} + \frac{3 x + 15}{7 ( 7 x ^{2} - 3 )}

3 + x - 4

What is the domain of f(x)=sqrt(12x^2-5x-150) ?
The domain of f(x)=sqrt(12x^2-5x-150) is x<=-10/3 \lor x>= 15/4