What is the domain of f(x)=sqrt(-4x^2+12) ?

The domain of f(x)=sqrt(-4x^2+12) is -sqrt(3)<= x<= sqrt(3)

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domain of f(x)=sqrt(-4x^2+12)

domain

f (x) = - 4 x^{2} + 12

- 3 \leq x \leq 3

Interval Notation

[- 3, 3]

Solution steps

Find non-negative values for radicals:

- 3 \leq x \leq 3

The function domain

- 3 \leq x \leq 3

f (x) = - 2 x + 10

y - 4 = \frac{4}{5} (x - 1)

f (x) = x^{2} - 25

f (x) = 3 x^{4} + 4 x^{3} + 6 x^{2} - 4

f (x) = 0.2 (x - 1)^{2} - 4

What is the domain of f(x)=sqrt(-4x^2+12) ?
The domain of f(x)=sqrt(-4x^2+12) is -sqrt(3)<= x<= sqrt(3)