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

  • Thread starter Thread starter mimitka
  • Start date Start date
  • Tags Tags
    Domain
AI Thread Summary
The function f(x) = x^(5/3) - 5x^(2/3) has a domain that is not all real numbers due to the nature of the exponents involved. Specifically, the calculator indicates that x cannot be negative because the expression involves roots that are undefined for negative inputs. The discussion highlights confusion around the behavior of graphing calculators, particularly the TI-83, which may misrepresent the function's domain. Proper factorization reveals that the function is defined for all non-negative values of x, leading to the conclusion that the domain is [0, ∞). Understanding the properties of exponents is crucial for determining the correct domain of the function.
mimitka
Messages
3
Reaction score
0
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!
 
Physics news on Phys.org
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?
 
AdmRucinski22 said:
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?
 

Similar threads

Find the domain and the range of ##f-3g##
Replies
3
Views
1K
Finding the domain (and range) of a logarithmic function
Replies
11
Views
3K
F(x)=(x+1)^0.5, h(x)=x^2+3, what is the range of f(h(x))?
Replies
4
Views
1K
How to find the range of a function with square roots?
Replies
23
Views
2K
Finding the domain of a composite function
Replies
10
Views
2K
To find the range of a given ##\sin## function
Replies
15
Views
2K
Is it possible to find the range of this function?
Replies
22
Views
1K
How to identify the type of function?
Replies
23
Views
3K
Need help in manipulating rational absolute value inequalities
Replies
11
Views
2K
What is the domain of f(f(x)) for f(x)= x/(1+x)?
Replies
3
Views
1K

Hot Threads

Recent Insights

Back
Top