What is the domain of f(x)=sqrt((x-2)(x+1)(1-3x)) ?

The domain of f(x)=sqrt((x-2)(x+1)(1-3x)) is x<=-1\lor 1/3 <= x<= 2

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domain of f(x)=sqrt((x-2)(x+1)(1-3x))

domain

f (x) = (x - 2) (x + 1) (1 - 3 x)

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

Interval Notation

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

Solution steps

Find non-negative values for radicals:

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

The function domain

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

\frac{x + 5}{2 x - 5}

f (x) = \frac{2 x + 3}{4 x - 7}

f (x) = ln (\frac{1}{x - 3})

- 4 x^{2} - 20 x - 21

g (x) = x - 4

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