Mathematics, 30.10.2020 02:30, saniyaf125

Why is the domain of f(x) = cube root of x NOT restricted while the domain of g(x) = sqrt(x) is? We cannot take the square root of any numbers except for those that are perfect squares, such as 0, 1, 4, 9, 16, etc.
The cube root of a negative number is still a real number, but the square root of a negative number is not real.
It is easier to take the square root of a number than it is to take the cube root of a number.
This is not true. The domains of both functions are restricted since the cube root and square root of negative numbers are both not real.

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Why is the domain of f(x) = cube root of x NOT restricted while the domain of g(x) = sqrt(x) is? We...