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

• mimitka
In summary, the function f(x) = x^(5/3) - 5x^(2/3) has a domain of all real numbers, but some graphing calculators may show that negative values are undefined due to the way they calculate exponents. Additionally, when factoring the function, the x-term should not be negative. The function F(x) = (2/5)^x has a domain of all real numbers, not just negative numbers.
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!

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??

mimitka said:
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!

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:
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).

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

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...)

The domain of F(x) = (2/5)^x is all negative numbers, right?

The domain of F(x) = (2/5)^x is all negative numbers, right?

No, it's not just "all negative numbers." You may want to look at its graph and see. Remember that the domain is the set of all INPUT values (usually x-values) of a function.

Also, may I ask why did you reply to a 2-year old thread with a new question?

## 1. What is the domain of the given function?

The domain of a function is the set of all possible input values for which the function is defined. In this case, the given function is a polynomial, which is defined for all real numbers. Therefore, the domain of f(x) = x5/3 - 5x2/3 is all real numbers.

## 2. Can the domain of a function be negative numbers?

Yes, the domain of a function can include negative numbers. In this case, the given function is defined for all real numbers, including negative numbers.

## 3. Are there any restrictions on the domain of this function?

No, there are no restrictions on the domain of this function. As mentioned before, the domain is all real numbers.

## 4. How can I determine the domain of a function algebraically?

To determine the domain of a function algebraically, you need to look for any values of the independent variable (x) that would make the function undefined. In this case, since the given function is a polynomial, it is defined for all real numbers, so there are no values that would make it undefined.

## 5. Can the domain of a function change depending on the given equation?

Yes, the domain of a function can change depending on the given equation. Each function has its own unique domain, and it can vary depending on the type of function and the restrictions on the independent variable. In this case, the domain of the given function is all real numbers, but this may not be the case for other functions.

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