# Critical Numbers

• Nov 20th 2009, 06:56 PM
rawkstar
Critical Numbers

find all critical numbers for the function f(x)=(9-x^2)^(3/5)

First thing i did was to find the derivative which i got -6x/[5(9-x^2)^(2/5)]

The next thing was to find when f'(x)=0, so i set the derivative equal to 0 to get x=0

I then checked to see when the derivative is undefined. To do this i set the denominator of the derivative [5(9-x^2)^(2/5)] to zero and got x=3 and x=-3.

are x=0,-3,3 right? I graphed the original function on my calculator and can't find anything specific about these points. also, i used the derivative function on my calculator to get the derivatives at these points, i thought that i'm not supposed to be able to do this since the derivative is supposed to be undefined at x=3 and x=-3
• Nov 20th 2009, 07:14 PM
ANDS!
The original function you have is defined over all real numbers. There is no need to check the domain of the derivative.
• Nov 20th 2009, 07:25 PM
rawkstar
how do you know that it is defined over all real numbers, i thought it would be undefined when the denominator equals 0
• Nov 20th 2009, 07:38 PM
Progenitor12
rawkstar,

I think you're right. The derivative is 0 when x=0 and undefined when x={-3,3}.

I graphed the function and it appears to have vertical asymptotes at x={-3,3} and it passes through 0 at x=0.
• Nov 20th 2009, 07:46 PM
rawkstar
the problem with mine is when i graph it, i am not getting the asymptotes at x=3 and x=-3
• Nov 20th 2009, 07:53 PM
Progenitor12
• Nov 20th 2009, 07:58 PM
rawkstar
ok i had a typo with my calculator input
i see it now, thank you
• Nov 20th 2009, 08:03 PM
ANDS!
Quote:

Originally Posted by rawkstar
how do you know that it is defined over all real numbers, i thought it would be undefined when the denominator equals 0

F(X) is a composition of functions. Each of which is defined over the reals. The derivative is just the derivative.
• Nov 20th 2009, 08:11 PM
Progenitor12
Quote:

Originally Posted by ANDS!
F(X) is a composition of functions. Each of which is defined over the reals. The derivative is just the derivative.

The function is undefined when the denominator is 0, that is, when $9-x^{2}=0$.

Even if two functions are defined for all x, their composed ratio is not *necessarily* defined for all x. It is undefined when the denominator is 0, just like any other fraction.
• Nov 20th 2009, 08:19 PM
ANDS!
Quote:

Originally Posted by Progenitor12
The function is undefined when the denominator is 0, that is, when $9-x^{2}=0$.

Even if two functions are defined for all x, their composed ratio is not *necessarily* defined for all x. It is undefined when the denominator is 0, just like any other fraction.

The original function is most definitely defined over the reals, regardless of the domain of it's derivative.

The behavior of the derivative (including its domain) will help describe the behavior of the original function, but the two are not necessarily linked as far as domain is concerned.