What is the domain of f(x)=sqrt(arcsin(3x-4x^3)) ?

The domain of f(x)=sqrt(arcsin(3x-4x^3)) is -1<= x<=-(sqrt(3))/2 \lor 0<= x<= (sqrt(3))/2

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domain of f(x)=sqrt(arcsin(3x-4x^3))

domain

f (x) = arcsin (3 x - 4 x^{3})

- 1 \leq x \leq - \frac{3}{2} or 0 \leq x \leq \frac{3}{2}

Interval Notation

[- 1, - \frac{3}{2}] \cup [0, \frac{3}{2}]

Solution steps

Find non-negative values for radicals:

- 1 \leq x \leq - \frac{3}{2} or 0 \leq x \leq \frac{3}{2}

The function domain

- 1 \leq x \leq - \frac{3}{2} or 0 \leq x \leq \frac{3}{2}

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

f (x) = \frac{x}{9} + 1

f (x) = 4 x ∣ x ∣

f (x) = \frac{4 + x}{4}

y = \frac{4 x + 2}{2}

What is the domain of f(x)=sqrt(arcsin(3x-4x^3)) ?
The domain of f(x)=sqrt(arcsin(3x-4x^3)) is -1<= x<=-(sqrt(3))/2 \lor 0<= x<= (sqrt(3))/2