Critical Point when f is undefined

In summary: 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.You should always multiply by the denominator.
  • #1
Cosmophile
111
2

Homework Statement


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

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

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

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

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 [tex] 1- \sqrt [3]x = 0 [/tex] leaving x = 1 as my only CP? Thanks!
 
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  • #2
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.
 
  • #3
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.
 
  • #4
Orodruin said:
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.
 
  • #5
Your quoted interval is [-1,1]. What is ##f(-1)##?
 
  • #6
Orodruin said:
Your quoted interval is [-1,1]. What is ##f(-1)##?

5, unless I'm horribly mistaken somewhere.
 
  • #7
Is ##(-1)^{2/3} = 1##?
 
  • #8
Rational powers can be well defined on negative Real numbers as long as the denominator is odd.

##(-1)^{3/2}## would be problematic.
 
  • #9
Orodruin said:
Is ##(-1)^{2/3} = 1##?

Yes.

PeroK said:
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.
 
  • #10
Cosmophile said:
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!
 
  • #11
Y
Cosmophile said:
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
 
  • #12
Orodruin said:
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.
 
  • #13
@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.
 
  • #14
Mark44 said:
@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
 
  • #15
PeroK said:
##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!
 
  • #16
PeroK said:
##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.
 
  • #17
Orodruin said:
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.
 
  • #18
SammyS said:
I believe all this discussion about branches, etc. has just confused OP.
You are probably correct. Let us drop that part of the discussion.
 
  • #19
Cosmophile said:
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.
 
  • #20
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.
 
  • #21
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.
 
  • #22
PeroK said:
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.
 

1. What is a critical point when f is undefined?

A critical point is a point on a function where the derivative is equal to zero or undefined. When f is undefined, it means that the function does not have a well-defined value at that particular point.

2. Why is a critical point important?

Critical points are important because they can help us identify maximum and minimum values of a function. They also give us information about the behavior and shape of a function.

3. How do you find the critical points when f is undefined?

To find the critical points when f is undefined, we need to take the derivative of the function and then set it equal to zero. This will give us the x-values of the critical points. However, if the derivative is undefined at a certain point, we need to further investigate to determine if it is a critical point or not.

4. Can a function have more than one critical point when f is undefined?

Yes, a function can have multiple critical points when f is undefined. This can happen when the derivative of the function is undefined at more than one point.

5. How do critical points when f is undefined differ from regular critical points?

Critical points when f is undefined differ from regular critical points because they do not have a well-defined value. This means that they may not necessarily be maximum or minimum points on the function, and further investigation is needed to determine their nature.

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