Grapher and odd roots

[freedrinks]

Any one notice that grapher will not graph the negative inputs for any function f(x)=x^a/b if b is odd and a>1. Any one know why this is?

 

[MisterMe]

Any one notice that grapher will not graph the negative inputs for any function f(x)=x^a/b if b is odd and a>1. Any one know why this is?

You did not accurately describe the "problem." However, the answer is simple. Grapher is a simple program. It does not intelligently handle imaginary or complex values of functions like yours.

 

[raindancer]

I'm coming onto this late, but it's a perfectly reasonable question. y=x^(2/3) does not involve any imaginary or complex numbers, and is equal to y=(x^(1/3))^2, which it will graph.

I think it's an interesting blip of stupidity in an otherwise semi-smart app.