What is the range of the function y = startroot x 5 endroot?
The Correct Answer and Explanation is :
The correct answer is: the range of the function ( y = \sqrt[5]{x} ) is all real numbers
The given function is ( y = \sqrt[5]{x} ), which represents the fifth root of ( x ). Let’s explore the domain, range, and properties of this function.
Range of the Function ( y = \sqrt[5]{x} ):
The range refers to all possible output values (or ( y )-values) of the function. For the fifth root function, ( y = \sqrt[5]{x} ), the behavior is as follows:
- Odd Root Behavior: Since the function involves taking the fifth root (an odd root) of ( x ), it can handle both positive and negative values of ( x ). Unlike even roots (like square roots), odd roots of negative numbers are defined. For example, ( \sqrt[5]{-32} = -2 ), because ( (-2)^5 = -32 ).
- All Real Numbers: For any real number ( x ), the fifth root is also a real number. As a result, ( y = \sqrt[5]{x} ) can produce both positive and negative outputs. Specifically:
- If ( x ) is positive, ( y ) is positive.
- If ( x ) is negative, ( y ) is negative.
- If ( x = 0 ), then ( y = 0 ).
Therefore, the range of the function ( y = \sqrt[5]{x} ) is all real numbers, which is mathematically expressed as:
[
\text{Range: } (-\infty, \infty)
]
Explanation:
The key reason the range of this function is all real numbers lies in the fact that the fifth root of any real number (whether positive, negative, or zero) always exists in the real number system. The function is continuous and unbounded, meaning that as ( x ) increases or decreases without bound, ( y ) also increases or decreases without bound. This distinguishes it from even root functions, which are restricted to non-negative ( x )-values because the square root of a negative number does not exist in the set of real numbers. Hence, the range of ( y = \sqrt[5]{x} ) extends from ( -\infty ) to ( \infty ).