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Take the following:

$$f(x) = x^{6/4}$$

The domain of this function is all real numbers. This function can be simplified to:

$$f(x) = x^{3/2}$$

The domain of this function is all real numbers greater than or equal to 0. Why is this true? Why does simplifying the function change its domain?

Jack Humphries
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  • I do not agree that the natural domain of the function given by the first expression is the set of all reals. – André Nicolas Oct 06 '13 at 02:55
  • @AndréNicolas Could you explain why? – Jack Humphries Oct 06 '13 at 03:04
  • What happens when you try to evaluate the function at $-1$? – tylerc0816 Oct 06 '13 at 03:06
  • Because we want our function to be continuous as a function of the exponent. For negative $x$, and irrational exponent near $6/4$, the function is not defined. Also, $6/4=3/2$. We do not want definedness to depend on the form we use. Also, for $(-5)^{6/4}$, what if we decided to take the fourth root first? – André Nicolas Oct 06 '13 at 03:10

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The problem is that $$ x^{(m/n)} = (x^m)^{1/n} $$ is only valid if $x$ is positive. So in other words, while of course it is true that $x^{(6/4)} = x^{(3/2)}$, you seem to be interpreting this to mean $(x^6)^{1/4} = (x^3)^{1/2}$, which doesn't follow.

Caleb Stanford
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