# What's wrong with my Mac Grapher?

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."

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Hello, I think you are graphing wrong function if your question is correct. it should be $y=x^{\frac{-2}{3}}$ and not $y=x^{\frac{2}{3}}$ – Paul Dirac Nov 28 '12 at 16:25
In fact, Grapher assumes the domain of $x^{1/3}x^{1/3}$ is all real numbers, but not $x^{2/3}$. – user29743 Nov 28 '12 at 16:26
@PaulDirac I think in between posting his question and his pictures, OP recognized that Grapher has this odd feature whether or not the exponent is negative. – user29743 Nov 28 '12 at 16:26
@countinghaus yes you're right, I just thought I would point it out. I don't think there's something wrong about your Grapher, I plotted same two functions on Mathematica, and got same results, but I don't know why. – Paul Dirac Nov 28 '12 at 16:31
@Paul Dirac Hello，thanks for the help,pictures edited now! – Ave Maleficum Nov 28 '12 at 16:48

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).

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Thank you,god Sir.Then how can I make Grapher(my plotting program) draw pictures on the whole X axis? – Ave Maleficum Nov 30 '12 at 9:21
Apply the power to the absolute value of $x$, if that makes sense. – Yuval Filmus Dec 3 '12 at 8:30
But that change the function. – Ave Maleficum Dec 7 '12 at 12:11

The grapher is incorrect here. The way the grapher is deciding domains is incorrect. x^(-2/3) should graph on both sides. If x^(-3/2) was your graph then it should not exist on the negatives. Note: Mathematica, Maple, and Grapher makes these mistakes. But TI-instruments do not. Even there lowly TI-83 which just plugs and chugs every number graphs this function correctly.

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