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Why is the graph of $f(x) = (x^2)^\frac{1}{6}$ not the same graph as $f(x) = x^\frac{1}{3}$?

Shouldn't they be the same because when you apply the exponent rules to the first equation, you get the same result as the second equation?

By the way, I am an Algebra 2 student in High School.

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    $\begingroup$ Hint-When you square something the negative disappears. $\endgroup$
    – KingLogic
    Commented Nov 25, 2020 at 2:43
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    $\begingroup$ Exponent rules don't work with negative bases and fractional exponents; cf. this question $\endgroup$ Commented Nov 25, 2020 at 2:48
  • $\begingroup$ The short answer is that $(x^p)^q=x^{pq}$ only for $x>0$ when $p$ and $q$ are rational numbers. $\endgroup$
    – Clayton
    Commented Nov 25, 2020 at 2:50
  • $\begingroup$ See also math.stackexchange.com/questions/2085268/does-i4-equal-1/…, and in particular the second-to-last paragraph. $\endgroup$
    – mweiss
    Commented Nov 25, 2020 at 3:00
  • $\begingroup$ Genuine question: is $-1^ \frac13$ where the operation " ^ $^ \frac13$ " is acting upon real numbers only and not in the complex realm, generally considered to be equal to $-1$, or is it undefined? Or do different texts have different conventions? Ah, J. W. Tanner has linked to the answer to my question... $\endgroup$ Commented Nov 25, 2020 at 3:07

1 Answer 1

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When you square the quantity $x$, the negative sign disappears, and it doesn't come back if you raise it to the $\frac{1}{6}$ power. If you just take $\frac{1}{3}$ power, then the negative stays negative.

$((-1)^2)^\frac{1}{6}=1^\frac{1}{6}=1$, while $(-1)^\frac{1}{3}=-1$

Therefore, what you see on the first graph is the absolute value of the second graph.

This hyperlink contains a graph of the two equations. Note that when $x<0$ the red solid line is the blue dotted line reflected about the x-axis.

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