subject
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.

ansver
Answers: 2

Another question on Mathematics

question
Mathematics, 21.06.2019 14:00
Bruce is getting materials for a chemistry experiment his teacher gives him a container that has 0.25 liter of liquid in it.bruce need to use 0.4 of this liquid for the experiment. how much liquid will bruce use?
Answers: 3
question
Mathematics, 21.06.2019 19:00
Find the length of the diagonal of the rectangle. round your answer to the nearest tenth. || | | 8 m | | | | 11 m
Answers: 2
question
Mathematics, 21.06.2019 21:30
Look at triangle wxy what is the length (in centimeters) of the side wy of the triangle?
Answers: 1
question
Mathematics, 21.06.2019 22:40
The value of x in this system of equations is 1. 3x + y = 9 y = –4x + 10 substitute the value of y in the first equation: combine like terms: apply the subtraction property of equality: apply the division property of equality: 3x + (–4x + 10) = 9 –x + 10 = 9 –x = –1 x = 1 what is the value of y?
Answers: 1
You know the right answer?
Why is the domain of f(x) = cube root of x NOT restricted while the domain of g(x) = sqrt(x) is? We...
Questions
Questions on the website: 13722367