The original function you have is defined over all real numbers. There is no need to check the domain of the derivative.
so I tried doing this problem but I'm having difficulty, please help
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
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.
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.