Critical Point when f is undefined

[Cosmophile]

Homework Statement

I am being asked to find the absolute extrema of the function

f(x) = 3x^{2/3}- 2x, [-1,1]

2. The attempt at a solution
I have taken the derivative and simplified it to:

\frac {1- \sqrt [3] x}{\sqrt [3]x} = 0

Here is my dilemma: 0 lies in the interval, and I know that f'(x) is undefined at x=0. Does this make x=0 a critical point? Is it wrong to multiply both sides by the denominator and simply be left with 1- \sqrt [3]x = 0 leaving x = 1 as my only CP? Thanks!

 

[Orodruin]

What branch of $x^{2/3}$ are you using for negative x? Regardless, if f' is not defined you will simply have to check those points explicitly.

 

[MarcusAgrippa]

Answers: 1. Ask: what is the definition of a critical point? Then check whether "undefined" is classified as a "critical point".
2. No, it is not wrong. And 1 is a critical point.

 

[Cosmophile]

What branch of $x^{2/3}$ are you using for negative x? Regardless, if f' is not defined you will simply have to check those points explicitly.

What do you mean what branch of $x^{2/3}$ am I using for negative x? Sorry to ask what is most likely a silly question.

 

[Orodruin]

Your quoted interval is [-1,1]. What is $f(-1)$?

 

[Cosmophile]

Your quoted interval is [-1,1]. What is $f(-1)$?

5, unless I'm horribly mistaken somewhere.

 

[Orodruin]

Is $(-1)^{2/3} = 1$?

 

[PeroK]

Rational powers can be well defined on negative Real numbers as long as the denominator is odd.

$(-1)^{3/2}$ would be problematic.

 

[Cosmophile]

Is $(-1)^{2/3} = 1$?

Yes.

Rational powers can be well defined on negative Real numbers as long as the denominator is odd.

$(-1)^{3/2}$ would be problematic.

I'm pretty sure I understand what you're saying. My question here was whether or not I should keep the denominator when I get a derivative in a form such that, say, x is in the denominator, making the derivative undefined when x=0. OR, should I always make it a point to multiply both sides by the denominator in order to only have a numerater = 0.

 

[Orodruin]

Yes.

This is true only for a very particular interpretation, which also breaks the relation $a = b^c \ \Rightarrow \ b = a^{1/c}$. Generally, the solution to an equation involving $x$ must be taken with that in mind. How do you interpret $x^{1/3}$ in your derivative? It must be taken with the same interpretation as you use for the function!

 

[PeroK]

Y

Yes.
I'm pretty sure I understand what you're saying. My question here was whether or not I should keep the denominator when I get a derivative in a form such that, say, x is in the denominator, making the derivative undefined when x=0. OR, should I always make it a point to multiply both sides by the denominator in order to only have a numerater = 0.

Drawing a graph of f(x) will answer your question about whether 0 is a critical point.

I can't see that it matters about having a denominator.

a/b = 0 iff a = 0, where b not = 0

 

[PeroK]

This is true only for a very particular interpretation, which also breaks the relation $a = b^c \ \Rightarrow \ b = a^{1/c}$. Generally, the solution to an equation involving $x$ must be taken with that in mind. How do you interpret $x^{1/3}$ in your derivative? It must be taken with the same interpretation as you use for the function!

$x^{1/3}$ and $x^{2/3}$ look like well defined real functions to me. Differentiable everywhere except 0. There's no ambiguity in the function values or derivatives that I can see.

$x^{2/3}$ is not invertible unless you restrict it to non-negative x, but that's not relevant here.

 

[Mark44]

@Cosmophile, your thread title ('Critical Point when f is undefined') might be an indication of your misunderstanding. Per wikipedia, a critical point is a point in the domain of f at which f' = 0 or is undefined. For the function in this problem, f(x) is defined for all real numbers. It is true that f' is not defined at x = 0, but the function itself is defined at x = 0.

 

[Cosmophile]

@Cosmophile, your thread title ('Critical Point when f is undefined') might be an indication of your misunderstanding. Per wikipedia, a critical point is a point in the domain of f at which f' = 0 or is undefined. For the function in this problem, f(x) is defined for all real numbers. It is true that f' is not defined at x = 0, but the function itself is defined at x = 0.

Nope, that's just a silly typo on my end that I didn't catch. Thanks! Haha

 

[Cosmophile]

$x^{1/3}$ and $x^{2/3}$ look like well defined real functions to me. Differentiable everywhere except 0. There's no ambiguity in the function values or derivatives that I can see.

$x^{2/3}$ is not invertible unless you restrict it to non-negative x, but that's not relevant here.

I thought they were well-defined, too.

So, I do need to include 0 as one of my critical numbers because f'(x) is undefined at x = 0. That makes sense to me. I just hadn't remembered the entire definition of a critical point, so I thought to just multiply both sides by the denominator.

Thank you, everyone!

 

[Orodruin]

$x^{1/3}$ and $x^{2/3}$ look like well defined real functions to me. Differentiable everywhere except 0. There's no ambiguity in the function values or derivatives that I can see.

$x^{2/3}$ is not invertible unless you restrict it to non-negative x, but that's not relevant here.

Did you ever try feeding it into Mathematica or Wolfram alpha? The analytic continuation from the real axis certainly is not a real function on the same branch as it is real on the positive axis. In order to make it real on the negative axis you need to pick a different branch.

 

[SammyS]

Did you ever try feeding it into Mathematica or Wolfram alpha? The analytic continuation from the real axis certainly is not a real function on the same branch as it is real on the positive axis. In order to make it real on the negative axis you need to pick a different branch.

In the problem at hand, and at the level it's presented It's a real function over a real domain.

I believe all this discussion about branches, etc. has just confused OP.

 

[Orodruin]

I believe all this discussion about branches, etc. has just confused OP.

You are probably correct. Let us drop that part of the discussion.

 

[Orodruin]

I thought they were well-defined, too.

So, I do need to include 0 as one of my critical numbers because f'(x) is undefined at x = 0. That makes sense to me. I just hadn't remembered the entire definition of a critical point, so I thought to just multiply both sides by the denominator.

Thank you, everyone!

Yes, you need to include it. The function can have an extremum where the derivative is not defined.

 

[Cosmophile]

Yep yep, solved! And the discussion of the branches didn't necessarily confuse me; in fact, it all sounds extremely interesting, and I'd love to know more about it. I'm going through Serge Lang's text and am using a different textbook's practice problems so that I can check answers (my copy of Lang's text is a PDF, and skipping to answers is tedious), and I was just eager to figure this out so I can move on. I appreciate the insight - even the insights that weren't necessarily tuned for my question.

 

[PeroK]

In real analysis (the basis of calculus) a function is usually defined to have only one value at each point. If you take something like $y = \pm \sqrt{x}$ then that is not a function, as there are two values for y for each non-zero x.

There are two functions there:

$y = \sqrt{x}$ and $y= - \sqrt{x}$

You could alternatively describe y as a " multi-valued function" with two branches.

Note that in real analysis the square root function is not defined for negative x.

If you move on to complex analysis then multi- valued functions become more common. In particular, every real number has 2 square roots and 3 cube roots. And, in fact, every real number has one real cube root and two complex cube roots.

In the same way that you can have $\pm \sqrt{x}$ the complex cube root has three branches.

The debate above is whether defining

$y = x^{1/3}$

Where y is taken to be the real cube root of x gives a well-defined function for any real x, positive or negative.

 

[Ray Vickson]

In real analysis (the basis of calculus) a function is usually defined to have only one value at each point. If you take something like $y = \pm \sqrt{x}$ then that is not a function, as there are two values for y for each non-zero x.

There are two functions there:

$y = \sqrt{x}$ and $y= - \sqrt{x}$

You could alternatively describe y as a " multi-valued function" with two branches.

Note that in real analysis the square root function is not defined for negative x.

If you move on to complex analysis then multi- valued functions become more common. In particular, every real number has 2 square roots and 3 cube roots. And, in fact, every real number has one real cube root and two complex cube roots.

In the same way that you can have $\pm \sqrt{x}$ the complex cube root has three branches.

The debate above is whether defining

$y = x^{1/3}$

Where y is taken to be the real cube root of x gives a well-defined function for any real x, positive or negative.

To clarify still more, we can write $x^{1/3} = \text{sign}(x) \, |x|^{1/3}$ in this case. Of course, that makes $x^{2/3} = |x|^{2/3}$ in this problem.