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

[MSchott]

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?

 

[acabus]

I don't think that $a^{b/c} = \sqrt[c]{a^b}$ can be used when a is negative.

 

[MSchott]

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?

 

[Mark44]

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 x^2/3 to a different form, using e as the base rather than x.

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

=> e^ln(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.

 

[MSchott]

Mark 44 Thank you so much.