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Specifically, consider the example $\sqrt[4]{x^2}$. The answer of course would be $\sqrt{|x|}$ since the x is squared first. However if converted to the exponential fraction of $x^{2/4}$, you lose the information of which came first, so you could end up with the wrong answer in this case. In fact, one might simplify it to $x^{1/2}$ which would simply be $\sqrt{x}$ which is NOT equal to $\sqrt{|x|}$.

Is there a proper convention for dealing with this? Or is the takeaway that you should be careful when simplifying a function into a fractional exponent?

The reason this is important to me is because I'm tutoring a student taking college algebra and I want to be 100% correct in my explanation of this sort of problem. He encounters many problems where he is asked to find the domain of functions and if he's ever given this type of scenario I need to know what to tell him.

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They key is just to track the domain all the way through the problem. If the domain becomes restricted as you simplify, you incorporate an absolute value to deal with it.

$$ \sqrt[4]{x^2}, x\in\mathbb{R} $$ $$ = x^{\frac{2}{4}}, x\in\mathbb{R} $$ $$ = |x|^{\frac{1}{2}}, x\in\mathbb{R} $$ $$ = \sqrt{|x|}, x\in\mathbb{R} .$$

If you miss the absolute value in the third step above, your domain becomes $x\in[0,\infty)$, so you've artificially restricted it. That's how you know you need to include the absolute value.

Just an additional note on why this works - the problem comes from the simplification of the fraction, and if you're really careful about your order of operations, it works:

$$ \sqrt[4]{x^2} = \left(x^2\right)^{\frac{1}{4}} = \left(\left(x^2\right)^{\frac{1}{2}}\right)^{\frac{1}{2}} = |x|^{\frac{1}{2}} = \sqrt{|x|}.$$

If the order had been reversed, we'd have:

$$ \sqrt[4]{x}^2 = \left(x^{\frac{1}{4}}\right)^2 = \left(\left(x^{\frac{1}{2}}\right)^{\frac{1}{2}}\right)^2 = x^{\frac{1}{2}} = \sqrt{x}.$$

The domain throughout this reversed problem is $x\in [0,\infty)$.

The key to it is in the following two identities:

$$ \left(x^2\right)^{\frac{1}{2}} = |x|, x\in\mathbb{R} , $$ $$ \left(x^{\frac{1}{2}}\right)^2 = x, x\in [0,\infty) .$$

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