How Many Critical Numbers Does the Function (3x-x^3)^(1/3) Have?

[Qube]

Homework Statement

I'm trying to determine the number of critical numbers for the function (3x-x^3)^(1/3).

Homework Equations

Critical numbers are where the derivative of the function is = 0 or does not exist.

Critical numbers must also exist within the domain of the function.

The Attempt at a Solution

I'm getting five critical numbers. Wolfram Alpha, however, disagrees, saying there are only 2 critical numbers. Who's right?

http://www.wolframalpha.com/input/?i=+(3x-x^3)^(1/3)+critical+numbers

This is my work:

1/3(3x-x^3)^(-2/3)(3-3x^2) = 0

3 - 3x^2 = 0

x = ±1

If x = 0 that will zero the denominator in the term with the negative root, and f'(x) will subsequently fail to exist.

If x = ±√3 that will zero the overall derivative.

All these points do exist on f(x).

 

[Ray Vickson]

Homework Statement

I'm trying to determine the number of critical numbers for the function (3x-x^3)^(1/3).

Homework Equations

Critical numbers are where the derivative of the function is = 0 or does not exist.

Critical numbers must also exist within the domain of the function.

The Attempt at a Solution

I'm getting five critical numbers. Wolfram Alpha, however, disagrees, saying there are only 2 critical numbers. Who's right?

http://www.wolframalpha.com/input/?i=+(3x-x^3)^(1/3)+critical+numbers

This is my work:

1/3(3x-x^3)^(-2/3)(3-3x^2) = 0

3 - 3x^2 = 0

x = ±1

If x = 0 that will zero the denominator in the term with the negative root, and f'(x) will subsequently fail to exist.

If x = ±√3 that will zero the overall derivative.

All these points do exist on f(x).

According to your definition (which includes places where f is not differentiable), there should be 4 critical points: x = ± 1, x= 0 and x = -√3; the derivative = 0 at x = ± 1 and the derivative does not exist at x = 0 or x = -√3. You might argue about x = +√3, but in a sense the derivative *does* exist there, but just happens to equal -∞. Anyway, if you look at critical points as possible maximizing or minimizing points, the four I listed do fulfill that criterion, while the last one (x = √3) does not. Just draw a graph and see for yourself.

 

[Qube]

Wait, what other definitions are there?

Also I think existence refers to the derivative being a real number, not infinity.

 

[Ray Vickson]

Wait, what other definitions are there?

Also I think existence refers to the derivative being a real number, not infinity.

Some books and papers regard derivatives of +∞ or -∞ as "existing" in the extended real number system. Others do not. So, some books would say that $x = \pm \sqrt{3}$ and $x=0$ are _not_ critical points, because the derivative "exists" but is not zero at those points; that is, $+\infty$ and $-\infty$ are not zero. Other books and papers would say that those ARE critical points because the derivative does not exist there. (I made an error before; I was looking at the wrong function. The derivative really does exist in the extended real number system.) To summarize: for some authors, the number of critical points is 2; for other authors it is 5.

BTW: SOME sources define critical numbers to be those values of f where the derivative vanishes; they do not consider points where the derivative fails to exist as a finite number. Those sources would say there are just two critical points, and there would be no need to get fancy with $+\infty$ or $-\infty$.

I think is fair to say that you should go with whatever definition your textbook and/or course notes prescribe.