Is square root x 2 exponent = x an identity (true for all nonnegative values of x)?

Answer 1

sqrt (x^2) =x if x>=0, so
yes, you are correct .

Answer 2

So you want to know if sqrt(x^2) = x is an identity for all nonnegative values of x.

The answer is yes. In actuality,

sqrt(x^2) = |x| (absolute value of x), but if you restrict this to nonnegative values, then for nonnegative values, |x| = x, so

sqrt(x^2) = x

An example of showing how this is false for negative numbers is if we let x = -1, then the left hand side is equal to sqrt([-1]^2) = sqrt(1) = 1, BUT, the right hand side is equal to -1. That is

sqrt([-1]^2) = (-1)
sqrt(1) = -1
1 = -1
Which is a false statement.

Answer 3

the sq. root of x to the second one ability is x. via the undeniable fact that is an same as declaring x to the flexibility of two/2. the dice root of x to the flexibility of three, will be x. any volume rooted to an same ability because the exponent of the bottom will go away you with merely the bottom, yet when the basis volume and the exponent are diverse then that is yet another tale all jointly, operating example, once you've the 7 root of x to the flexibility of four, that's an same as x to the 4/7 ability. if so, x will be taken to the flexibility of four, then rooted to the flexibility of seven.

Answer 4

it is true for all values of x