Why is the domain of this rational function restricted to x > 0?

[raxAdaam]

When I plot

y = 3x^{\frac{4}{3}}-\frac{3}{32}x^{\frac{2}{3}},

I get: https://dl.dropbox.com/u/5653705/pfgraph1.png


but I don't see any reason for the domain to be restricted to x > 0. There is only a cubic root, which is well defined for negative numbers ... I've tried on a few different programs & always get the same thing, can anyone explain why? Is the domain actually restricted to positive x?

Thanks in advance,


Rax

 

[jbunniii]

A negative number has three distinct cube roots. For example, all three of the following are cube roots of $-1$.

the real root:
$$-1$$
the principal root:
$$e^{i\pi/3} = \cos(\pi/3) + i\sin(\pi/3) = \frac{1}{2} + i\frac{\sqrt{3}}{2}$$
the conjugate of the principal root:
$$e^{-i\pi/3} = \cos(\pi/3) - i\sin(\pi/3) = \frac{1}{2} - i\frac{\sqrt{3}}{2}$$
Many calculators, particularly those that don't handle complex numbers, will give the $-1$ answer. Software packages that know about complex numbers will likely give the principal cube root, which is $e^{i\pi/3}$.

 

[Mark44]

The reason you're getting only the part of the graph for x ≥ 0 is that the software is very likely representing the x^4/3 and x^2/3 terms in terms of e to some power.

Looking at the first of these functions, we have

$$ x^{4/3} = (e^{\ln x})^{4/3} = e^{4/3 \cdot \ln x}$$

The above is not an identity, since the exponential expressions are defined only for x > 0, whereas the left side is defined for all real numbers.

 

[raxAdaam]

The reason you're getting only the part of the graph for x ≥ 0 is that the software is very likely representing the x^4/3 and x^2/3 terms in terms of e to some power.

Looking at the first of these functions, we have

$$ x^{4/3} = (e^{\ln x})^{4/3} = e^{4/3 \cdot \ln x}$$

The above is not an identity, since the exponential expressions are defined only for x > 0, whereas the left side is defined for all real numbers.

Interesting, and this tends to be the case only for rational exponents? So were I to factor the expression in terms of an integer exponent, I would get the entire graph? How come these programs don't account for this method? Wouldn't it be relatively simple to include the other half of the graph?


Thank you very much for the answer - very insightful!




Rax


PS I tried factoring & re-graphing, and sure enough go the full function! Very cool - thanks a ton, would never have figured that out on my own! Still don't understand why they wouldn't program around that - potentially misleading! Here's the graph rendered for y = \frac{3}{32}x \left(32x^{\frac{1}{3}} - x^{-\frac{1}{3}}\right), for anyone else interested:

https://dl.dropbox.com/u/5653705/pfgraph2.png

 

[Mark44]

Interesting, and this tends to be the case only for rational exponents?

Assuming you mean rational exponents vs. integer exponents, then yes. There's no problem graphing y = xⁿ, where x is an integer.

For rational exponents, things are different. If r = $\frac{m}{n}$, with the fraction reduced to lowest terms, we're going to run into problems if n is even, but not if n is odd. This means, for example, that y = x^2/3 is defined for all real numbers, while y = x^3/2 is defined only for x ≥ 0. I am restricting things to real-valued functions of real variables. I am excluding complex numbers from consideration.

Many graphing packages would use the scheme I described earlier to graph these functions, so for both, only the parts of the graph for which x > 0 would be shown.

So were I to factor the expression in terms of an integer exponent, I would get the entire graph? How come these programs don't account for this method? Wouldn't it be relatively simple to include the other half of the graph?

I don't think it would be all that simple, since the rational exponent would need to be reduced to lowest terms, and then decide whether the denominator was even or odd. At any rate, it's very simple to convert to an exponential form before graphing.

Thank you very much for the answer - very insightful!




Rax


PS I tried factoring & re-graphing, and sure enough go the full function! Very cool - thanks a ton, would never have figured that out on my own! Still don't understand why they wouldn't program around that - potentially misleading! Here's the graph rendered for y = \frac{3}{32}x \left(32x^{\frac{1}{3}} - x^{-\frac{1}{3}}\right), for anyone else interested:

https://dl.dropbox.com/u/5653705/pfgraph2.png