Domain and Range of Linear Functions

Oct 13, 2025 by orovajewels.com 60 views

Introduction

In mathematics, functions are used to describe the relationship between variables. The domain of a function is the set of all possible input values for which the function is defined, while the range is the set of all possible output values. In this article, we will explore the domain and range of the function g(x) = \sqrt[5]{x+2} - 1.

Domain of the Function

The domain of a function is the set of all possible input values for which the function is defined. In the case of the function g(x) = \sqrt[5]{x+2} - 1, we need to consider the restrictions on the input values.

Radical Function: The function involves a radical, specifically a fifth root. The expression inside the radical, x+2, must be non-negative, as the fifth root of a negative number is undefined.
Domain Restriction: To ensure that the expression inside the radical is non-negative, we must have x+2 ≥ 0. Solving this inequality, we get x ≥ -2.

Therefore, the domain of the function g(x) = \sqrt[5]{x+2} - 1 is all real numbers x such that x ≥ -2.

Range of the Function

The range of a function is the set of all possible output values. In the case of the function g(x) = \sqrt[5]{x+2} - 1, we need to consider the restrictions on the output values.

Radical Function: The function involves a radical, specifically a fifth root. The expression inside the radical, x+2, is always non-negative, as x ≥ -2.
Range Restriction: As the expression inside the radical is non-negative, the fifth root of x+2 will always be non-negative. Therefore, the output of the function g(x) = \sqrt[5]{x+2} - 1 will always be non-negative.
Minimum Value: The minimum value of the function occurs when x = -2, which gives g(-2) = \sqrt[5]{-2+2} - 1 = -1.
Maximum Value: As x increases without bound, the value of g(x) = \sqrt[5]{x+2} - 1 will also increase without bound.

Therefore, the range of the function g(x) = \sqrt[5]{x+2} - 1 is all real numbers y such that y ≥ -1.

Conclusion

In conclusion, the domain of the function g(x) = \sqrt[5]{x+2} - 1 is all real numbers x such that x ≥ -2, while the range is all real numbers y such that y ≥ -1. Understanding the domain and range of a function is crucial in mathematics, as it helps us to determine the possible input and output values of the function.

References

[1] "Functions" by Khan Academy. Retrieved from https://www.khanacademy.org/math/algebra-functions/functions
[2] "Domain and Range" by Math Open Reference. Retrieved from https://www.mathopenref.com/domainrange.html

Introduction

In our previous article, we explored the domain and range of the function g(x) = \sqrt[5]{x+2} - 1. In this article, we will answer some frequently asked questions about the domain and range of this function.

Q&A

Q: What is the domain of the function g(x) = \sqrt[5]{x+2} - 1?

A: The domain of the function g(x) = \sqrt[5]{x+2} - 1 is all real numbers x such that x ≥ -2.

Q: Why is the domain of the function g(x) = \sqrt[5]{x+2} - 1 restricted to x ≥ -2?

A: The domain of the function g(x) = \sqrt[5]{x+2} - 1 is restricted to x ≥ -2 because the expression inside the radical, x+2, must be non-negative. If x+2 is negative, the fifth root of x+2 is undefined.

Q: What is the range of the function g(x) = \sqrt[5]{x+2} - 1?

A: The range of the function g(x) = \sqrt[5]{x+2} - 1 is all real numbers y such that y ≥ -1.

Q: Why is the range of the function g(x) = \sqrt[5]{x+2} - 1 restricted to y ≥ -1?

A: The range of the function g(x) = \sqrt[5]{x+2} - 1 is restricted to y ≥ -1 because the expression inside the radical, x+2, is always non-negative. Therefore, the fifth root of x+2 will always be non-negative, and the output of the function g(x) = \sqrt[5]{x+2} - 1 will always be non-negative.

Q: What is the minimum value of the function g(x) = \sqrt[5]{x+2} - 1?

A: The minimum value of the function g(x) = \sqrt[5]{x+2} - 1 occurs when x = -2, which gives g(-2) = \sqrt[5]{-2+2} - 1 = -1.

Q: What is the maximum value of the function g(x) = \sqrt[5]{x+2} - 1?

A: The maximum value of the function g(x) = \sqrt[5]{x+2} - 1 does not exist, as the value of g(x) will increase without bound as x increases without bound.

Q: How do I determine the domain and range of a function?

A: To determine the domain and range of a function, you need to consider the restrictions on the input and output values. For example, if a function involves a radical, you need to ensure that the expression inside the radical is non-negative. If a function involves a fraction, you need to ensure that the denominator is non-zero.

Conclusion

In conclusion, the domain and range of the function g(x) = \sqrt[5]{x+2} - 1 are all real numbers x such that x ≥ -2 and all real numbers y such that y ≥ -1, respectively. Understanding the domain and range of a function is crucial in mathematics, as it helps us to determine the possible input and output values of the function.

References

[1] "Functions" by Khan Academy. Retrieved from https://www.khanacademy.org/math/algebra-functions/functions
[2] "Domain and Range" by Math Open Reference. Retrieved from https://www.mathopenref.com/domainrange.html