Why is the domain of ax^(1/3) + b equal to all real numbers?

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In summary, the domain of the function ax^(1/3) + b is all real numbers because there are no restrictions on the input value x. Exceptions to this rule include when the coefficient a is equal to 0, or when x is negative and a is an even number. This domain is similar to other polynomial functions, but different from functions with specific rules for input values. The inclusion of the cube root ensures continuity for all values of x, including negative values. The domain can be limited by additional restrictions or conditions given for the function.
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mathdad
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Find the domain.

ax^(1/3) + b

I understand that a and b are constants here. I also know that x^(1/3) is equivalent to cube root{x}.

What I do not understand is why the answer is ALL REAL NUMBERS.
 
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  • #2
It's "all real numbers" because no matter what real value you give x the function returns another real number. Given that a function's domain is defined wherever the function gives a real number for a real number input, ax^(/13) + b has the set $\mathbb{R}$ as its domain, i.e. it is defined for all real x.
 
  • #3
Thanks. Good information.
 

FAQ: Why is the domain of ax^(1/3) + b equal to all real numbers?

Why is the domain of ax^(1/3) + b equal to all real numbers?

The domain of a mathematical function refers to the set of all possible input values for which the function is defined. In the case of the function ax^(1/3) + b, the domain is all real numbers because there are no restrictions on the input value x. In other words, any real number can be substituted for x and the function will still produce a real number as the output. This is because the cube root (^(1/3)) of any real number is also a real number, and adding or multiplying real numbers will always result in a real number.

Are there any exceptions to this rule?

Yes, there are a few exceptions to this rule. One exception is when the coefficient a is equal to 0, in which case the function becomes b, and the domain is restricted to only the value of b. Another exception is when the value of x is negative and a is an even number, in which case the function would result in a complex number. However, for the most part, the domain of ax^(1/3) + b is all real numbers.

How does the domain of ax^(1/3) + b compare to other functions?

The domain of ax^(1/3) + b is similar to many other polynomial functions, such as ax^2 + bx + c or ax^3 + bx^2 + cx + d. These functions also have a domain of all real numbers because there are no restrictions on the input values. However, functions such as sin(x) or log(x) have a restricted domain because they have specific rules for the input values.

Is there a specific reason why the function includes the cube root?

The inclusion of the cube root in the function ax^(1/3) + b is not arbitrary. The cube root is used to ensure that the function is continuous for all values of x, including negative values. For example, if the function was ax^(1/2) + b, it would not be defined for negative values of x because the square root of a negative number is not a real number. But by using the cube root, the function can be defined for all real numbers.

Can the domain of ax^(1/3) + b ever be limited?

Yes, the domain of ax^(1/3) + b can be limited if there are additional restrictions or conditions given for the function. For example, if the function was ax^(1/3) + b, but it is specified that x must be greater than or equal to 0, then the domain would be limited to only non-negative real numbers. Similarly, if the function was ax^(1/3) + b, but it is specified that x must be an integer, then the domain would be limited to only integer values.

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