What is the domain of f(x)=x^(2/3)?

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The domain of the function f(x) = x^(2/3) is all real numbers, as negative inputs yield positive outputs when squared, followed by taking the cube root. Some confusion arises from WolframAlpha suggesting the domain is all non-negative real numbers, which may stem from its interpretation of the function. The logarithmic transformation used in WolframAlpha implies that x must be greater than zero, excluding zero itself. However, since the cube root of a positive number is defined for all real values, the correct domain remains all real numbers. This highlights the importance of understanding different representations of the function.
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What is the domain of f(x)=x^(2/3)?

If I take a negative number and square it, the result is positive. For example, (-4)^2 is 16. The cube root is 16 is also a positive number. So I conclude that the domain of this function is "all real numbers". But in Wolfram the answer is "all non-negative real numbers". Where is my thinking going awry?
 
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I don't think that a^{b/c} = \sqrt[c]{a^b} can be used when a is negative.
 


acabus: I am still having difficulty understanding. If "b" is an even power doesn't the negative "a" under the square root sign become positive?
 


MSchott said:
What is the domain of f(x)=x^(2/3)?

If I take a negative number and square it, the result is positive. For example, (-4)^2 is 16. The cube root is 16 is also a positive number. So I conclude that the domain of this function is "all real numbers". But in Wolfram the answer is "all non-negative real numbers". Where is my thinking going awry?
You are correct - the domain of your function is all real numbers. The reason, I believe, that WolframAlpha gives a different domain is that it is converting x2/3 to a different form, using e as the base rather than x.

y = x2/3
=> ln(y) = (2/3) ln(x)

=> eln(y) = e(2/3)ln(x)
=> y = e(2/3)ln(x)

Now, x must be greater than 0. This is almost the same as what WA shows, except for zero.
 


Mark 44 Thank you so much.
 

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