What is the domain of f(x)=(sqrt(3x+4))(2x^2+3x) ?

The domain of f(x)=(sqrt(3x+4))(2x^2+3x) is x>=-4/3

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domain of f(x)=(sqrt(3x+4))(2x^2+3x)

domain

f (x) = (3 x + 4) (2 x^{2} + 3 x)

x \geq - \frac{4}{3}

Interval Notation

[- \frac{4}{3}, \infty)

Solution steps

Find non-negative values for radicals:

x \geq - \frac{4}{3}

The function domain

x \geq - \frac{4}{3}

f (x) = \frac{x + 3}{2} + \frac{x}{x - 3}

- \frac{1}{x + 3}

f (x) = \frac{x ( x + 8 )}{9 ( x - 5 )}

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

\frac{5 x - 7}{x + 8}

What is the domain of f(x)=(sqrt(3x+4))(2x^2+3x) ?
The domain of f(x)=(sqrt(3x+4))(2x^2+3x) is x>=-4/3