What is the domain of f(x)=-sqrt(4-x^2),0<= x<2 ?

The domain of f(x)=-sqrt(4-x^2),0<= x<2 is 0<= x<2

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domain of f(x)=-sqrt(4-x^2),0<= x<2

domain

f (x) = - 4 - x^{2}, 0 \leq x < 2

0 \leq x < 2

Interval Notation

[0, 2)

Solution steps

Set the domain to the given interval

0 \leq x < 2

Find non-negative values for radicals:

- 2 \leq x \leq 2

Join real regions with the domain:

- 2 \leq x \leq 2 and 0 \leq x < 2

0 \leq x < 2

y = - lo g_{2} (3 (x - 2)) + 4

25 - 5^{x}

f (x) = - 5 x + 3, - 2 \leq x \leq 4

f (x) = 12 x^{3} - 5 x^{2} - 11 x + 6

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

What is the domain of f(x)=-sqrt(4-x^2),0<= x<2 ?
The domain of f(x)=-sqrt(4-x^2),0<= x<2 is 0<= x<2