# Find domain of f(x)=x^(5/3)-5x^(2/3)

1. Feb 7, 2009

### mimitka

f(x)= x^(5/3) - 5x^(2/3)

Need to find the domain of the above function?
The graphing calculator shows x cannot be a negative number, and I don't see why not? Please help!

2. Feb 7, 2009

### d_b

Think of this function as (x^2 - x^3) and so if you have x to the power of anything will there be a value of x that function f(x) does not exists? ----if there is no value of x that f(x) does not exists then your domain is everything, but if there is a value(s) of x that f(x) does not exists(undefined) then your domain is everything except that value,

does this help??

3. Feb 8, 2009

My guess: Since the exponent is not an integer, the graphing calculator is using the formula xa = ea ln x, which is undefined for negative x. Really, its domain is the whole real number line (assuming you're working with real numbers), where we can let (-x)1/3 = -(x1/3).

Are you using a TI-83 or similar? The TI-83 (and probably any related calculator) is peculiar in that if you calculate, say, -11/2, it will give back -1, as if it's doing (-x)a = -(xa), for any a. Well, it's pretty wrong about that, but for some reason it does it differently (more correctly) when graphing.

Last edited: Feb 8, 2009
4. Feb 8, 2009

### OctavianIV

If you factorise f(x) as x^2/3(x^5/2 -5), the x term in the bracket cannot be negative (being the square root of x^5).

5. Feb 8, 2009

That's wrong. If you factor it correctly you get x2/3(x - 5).

6. Feb 8, 2009

### OctavianIV

Of course - sorry about that! I think we can just say the domain is all real x and the calculator is wrong (and so is my factorising...)

7. Jul 29, 2011