What is the domain of f(x)=sqrt(6x^2+7x-5) ?

The domain of f(x)=sqrt(6x^2+7x-5) is x = 1/2

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domain of f(x)=sqrt(6x^2+7x-5)

domain

f (x) = 6 x^{2} + 7 x - 5

x \leq - \frac{5}{3} or x \geq \frac{1}{2}

Interval Notation

(- \infty, - \frac{5}{3}] \cup [\frac{1}{2}, \infty)

Solution steps

Find non-negative values for radicals:

x \leq - \frac{5}{3} or x \geq \frac{1}{2}

The function domain

x \leq - \frac{5}{3} or x \geq \frac{1}{2}

f (x) = lo g_{3} (x - 1) + 0.239784

f (x) = \frac{1}{1 + x ^{2}}

- x^{2} + 3

\frac{1}{x ^{3} + 4 x}

f (x) = sin (x) + cos (x)

What is the domain of f(x)=sqrt(6x^2+7x-5) ?
The domain of f(x)=sqrt(6x^2+7x-5) is x<=-5/3 \lor x>= 1/2