Answered

Which function has a domain of all real numbers?


A$y=(2 x)^{\frac{1}{3}}-7$
B$y=-x^{\frac{1}{2}}+5$
C$y=-2(3 x)^{\frac{1}{6}}$
D$y=(x+2)^{\frac{1}{4}}$

Verified Answer :

  • Recall that a function \(f(x) = [g(x)]^{p/q}\) (with \(p/q\) in lowest terms) is defined for all real numbers if and only if \(q\) is an odd integer.
  • Evaluate each function's exponent denominator \(q\): \(\frac{1}{3}\) (\(q=3\)), \(\frac{1}{2}\) (\(q=2\)), \(\frac{1}{6}\) (\(q=6\)), and \(\frac{1}{4}\) (\(q=4\)).
  • Identify that only the first function has an odd denominator (\(q=3\)), making it the only function defined for all real numbers.
  • Final answer: \(\boxed{y=(2x)^{\frac{1}{3}}-7}\).

Explanation

  1. Analysis of Domain Conditions for Functions with Rational Exponents

To determine which function has a domain of all real numbers, we need to analyze the conditions under which each function is defined in the set of real numbers. A function of the form \(y = f(x)^n\), where the exponent \(n = \frac{p}{q}\) is in lowest terms with \(q > 0\), is defined for all real numbers if the operation \(\sqrt[q]{f(x)}\) is valid for any \(f(x) \in \mathbb{R}\). This occurs if and only if \(q\) is an odd integer. If \(q\) is an even integer, the domain is restricted because the root \(\sqrt[q]{z}\) is only defined for non-negative values of \(z\) (\(z \ge 0\)).

  1. Evaluating the Domain of Each Function

We examine the exponent \(n = \frac{p}{q}\) for each given function, ensuring the fraction is in lowest terms:

  1. For \(y = (2x)^{\frac{1}{3}} - 7\), the exponent is \(n = \frac{1}{3}\). Here, \(q = 3\), which is an odd integer. Therefore, \(\sqrt[3]{2x}\) is defined for all real numbers \(x\), meaning the function is defined for all \(x \in \mathbb{R}\).
  2. For \(y = -x^{\frac{1}{2}} + 5\), the exponent is \(n = \frac{1}{2}\). Here, \(q = 2\), which is an even integer. Thus, \(\sqrt{x}\) is defined only for \(x \ge 0\).
  3. For \(y = -2(3x)^{\frac{1}{6}}\), the exponent is \(n = \frac{1}{6}\). Here, \(q = 6\), which is an even integer. Thus, \(\sqrt[6]{3x}\) is defined only for \(3x \ge 0\), or \(x \ge 0\).
  4. For \(y = (x + 2)^{\frac{1}{4}}\), the exponent is \(n = \frac{1}{4}\). Here, \(q = 4\), which is an even integer. Thus, \(\sqrt[4]{x+2}\) is defined only for \(x + 2 \ge 0\), or \(x \ge -2\).
  1. Conclusion

Comparing the results from step 2, only the first function, \(y = (2x)^{\frac{1}{3}} - 7\), has an exponent with an odd denominator (\(q=3\)) in simplest form. Consequently, it is the only function in the list that is defined for any real number \(x\). The other functions are restricted to non-negative values of their inner arguments due to having even denominators (\(q=2, 6, 4\)) in their exponents. Therefore, the function \(y = (2x)^{\frac{1}{3}} - 7\) has a domain of all real numbers.

Examples

In engineering and data analysis, identifying the domain of a function is crucial for ensuring that calculations are valid. For instance, when modeling growth or decay processes using power functions, you must ensure that your model produces real-valued outputs for all possible inputs in your dataset. If a model involves even roots, inputs like negative numbers might lead to undefined values (imaginary results), which could crash a simulation or lead to incorrect analysis. Recognizing that odd-indexed roots allow for negative inputs helps analysts build more robust mathematical models for diverse scenarios.

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