I hope this question is not that stupid.
$y=x^{\frac{-2}{3}}$ is $y=\frac{1}{\sqrt[3]{x^{2}}} $right?
But when I type $y=x^{\frac{-2}{3}} $and $y=\frac{1}{\sqrt[3]{x^{2}}}$into Mac Grapher,I got different pictures.
Since my reputation is not over 10,I can't post them,but it's very easy to do that.$y=x^{\frac{-2}{3}}$ did not show on negative direction of X-axis,while $y=\frac{1}{\sqrt[3]{x^{2}}}$ showed on all X-axis.
EDIT:Thank you everyone,I can post pictures now
or more examples
and in some case they are same
Here are some polynomial examples. Ther're also equal
Another set
And I did some research ,you can see Mac Grapher only show function with integer exponents
Change 3 to 5
I could go on,but you get the ideas.
Why did this happen,is there something wrong with my precalculus skill or Mac Grapher?
Update：Some guy in Apple forum said"The functions are not the same. In the first case, you're raising a real number to a fractional power. That's ambiguous for negative arguments, so the program only graphs it for non-negative arguments. In the second case, you're raising a non-negative number to a fractional power."
Best Answer
Integer powers $\ldots,x^{-3},x^{-2},x^{-1},x^0,x^1,x^2,x^3,\ldots$ are always defined, since they can be defined using multiplication and multiplicative inverse (e.g. $x^3 = x \cdot x \cdot x$, $x^{-3} = 1/x^3$). When you extend the definition to rational numbers, the idea is to define $x^{a/b} = \sqrt[b](x^a)$. Yet every non-zero number has $b$ different complex roots! This is a problem since we want the rules of powers to hold. For example, we would like the equation $\sqrt{xy} = \sqrt{x}\sqrt{y}$ to hold. For that, we need a rule for choosing square roots (in general, $b$th roots) which is consistent in the sense that $\sqrt{xy} = \sqrt{x}\sqrt{y}$ is satisfied.
If $x$ is positive then we can always choose $\sqrt[b]{x}$ to be the unique $b$th square root. Otherwise, we run into a problem: there is no consistent way of choosing square roots across the complex plane so that $\sqrt{xy} = \sqrt{x}\sqrt{y}$! Technically, this is because the complex multivalued square root function has a branch point at the origin.
There are two ways to handle this problem. First, we could arbitrarily choose some value for $x^{a/b}$. While this value won't satisfy $(xy)^{a/b} = x^{a/b} y^{a/b}$, it could satisfy $x^{a/b+c/d} = x^{a/b} x^{c/d}$. This is implemented by choosing a branch cut of the complex logarithm function, usually the so-called principal branch. Second, we could just disallow $x^{a/b}$ when $x$ is not a positive number. The latter approach is the one taken by your plotting program.
You can ask, how come my program plots $x^{3/3}$ correctly even for negative $x$? The reason is that the program first simplifies $3/3 = 1$, realizes that $1$ is an integer, and so is able to compute $x^1$ for every value of $x$. In general, when the program encounters $x^y$, it check first whether $y$ is an integer; if so, the computation is always possible. Otherwise, it is possible only if $x$ is positive (or zero).