Critical points

[Loki123]

So I get that critical points are when f’(a) = 0 or f’(a) doesn't exist, but why is that? What exactly is a critical point, why do we get it?

 

[BigBeachBanana]

So I get that critical points are when f’(a) = 0 or f’(a) doesn't exist, but why is that? What exactly is a critical point, why do we get it?

Consider a simple example [imath]f(x)=x^2[/imath]. We know that it has a vertex at [imath]x=0[/imath], and it is also the minimum point of [imath]f(x)=x^2[/imath]. What is the slope at [imath]x=0?[/imath]

Consider a another example [imath]f(x)=(x-5)^2[/imath]. We know that it has a vertex at [imath]x=5[/imath], and it is also the minimum point. What's the slope at [imath]x=5?[/imath]

 

[Steven G]

You can only have a relative max or min where f'(x)=0, undefined or at and endpoint if given a closed interval.

 

[JeffM]

Critical points identify where a function potentially changes qualitatively.

If f’(x) = 0 when x = a, there may be a local minimum or maximum at a. If so, the function will be increasing on one side of a and decreasing on the other side of a.

If f’(x) does not exist when x = a, there is obvious change in the behavior of f(x) at a (if f(a) is even a real number).

They are critical because they identify points of change in behavior.

 

[Deleted member 4993]

If so, the function will be increasing on one side of a and decreasing on the other side of a.

I think that statement needs "revision".

 

[JeffM]

I think that statement needs "revision".

Oh great Khan, I shall revise it once I know how.

If f(x) is at a local maximum at a, then f(x) is increasing to the left of a and decreasing to the right; that means f(x) is increasing on one side of a and decreasing on the other.

If f(x) is at a local minimum at a, then f(x) is increasing to the right of a and decreasing to the left of a; that means f(x) is increasing on one side of a and decreasing on the other.

 

[Deleted member 4993]

Oh great Khan, I shall revise it once I know how.

If f(x) is at a local maximum at a, then f(x) is increasing to the left of a and decreasing to the right; that means f(x) is increasing on one side of a and decreasing on the other.

If f(x) is at a local minimum at a, then f(x) is increasing to the right of a and decreasing to the left of a; that means f(x) is increasing on one side of a and decreasing on the other.

You have, I think, neglected the situation where f"(x) = 0, we have a stationary point and the picture changes. I know this may not be pertinent at the level of OP..........but................so the statement

If f’(x) = 0 when x = a, there may be a local minimum or maximum at a

I know there is "if " and "may be" in that statement - I "feel" that statement should point to possibility of stationary point also. I remember while I was teaching mechanics, I had problem discussing "stable" equilibrium (minima and maxima situation being unstable equilibrium).

Anyway, this could be left to the future (but you know I can see future....)

 

[JeffM]

You have, I think, neglected the situation where f"(x) = 0, we have a stationary point and the picture changes. I know this may not be pertinent at the level of OP..........but................so the statement

I know there is "if " and "may be" in that statement - I "feel" that statement should point to possibility of stationary point also. I remember while I was teaching mechanics, I had problem discussing "stable" equilibrium (minima and maxima situation being unstable equilibrium).

Anyway, this could be left to the future (but you know I can see future....)

Ahh, I see. Yes, of course, I ignored important subtleties, very important subtleties I agree. I was trying to give an answer that was basic and avoided errors. If we start talking about dynamics and dx/dt, for example, a critical point that is a stable equilibrium is qualitatively different from an unstable equilibrium. Honestly, I was not trying to answer at that level of sophistication.

 

[Loki123]

Consider a simple example [imath]f(x)=x^2[/imath]. We know that it has a vertex at [imath]x=0[/imath], and it is also the minimum point of [imath]f(x)=x^2[/imath]. What is the slope at [imath]x=0?[/imath]

Consider a another example [imath]f(x)=(x-5)^2[/imath]. We know that it has a vertex at [imath]x=5[/imath], and it is also the minimum point. What's the slope at [imath]x=5?[/imath]

0? I know it's horizontal. What about when the derivative does not exist?

 

[Loki123]

Critical points identify where a function potentially changes qualitatively.

If f’(x) = 0 when x = a, there may be a local minimum or maximum at a. If so, the function will be increasing on one side of a and decreasing on the other side of a.

If f’(x) does not exist when x = a, there is obvious change in the behavior of f(x) at a (if f(a) is even a real number).

They are critical because they identify points of change in behavior.

I understand that. Why does a function need a critical point to change? Can't it just change without it?

 

[BigBeachBanana]

0? I know it's horizontal. What about when the derivative does not exist?

Right. Critical points are those points on a graph at which a line drawn tangent to the curve is horizontal (0) or vertical (undefined). There are three types of critical points- maximums, minimum, and points of inflection.

For the undefined case, consider:
(1) [imath]f(x)=\frac{x}{x-5}[/imath]
(2) [imath]f(x)=(x-1)^{2/3}[/imath]
Where is the derivative equal to 0? Where is the derivative undefined? Keep in mind the domain of the function when answering the questions.
I'll leave this for you as an exercise.

 

[Dr.Peterson]

Why does a function need a critical point to change [direction]? Can't it just change without it?

What about when the derivative does not exist?

The key idea is that a critical point is a point where the sign of the derivative may change. If the derivative is continuous, it can change only by passing through zero (though it doesn't have to change when it is zero). If the derivative is undefined, anything can happen because it is not continuous, so you need to check whether the sign changes. (That includes the function itself being undefined.)

So the definition of "critical point" is designed precisely to answer your first question above.

 

[JeffM]

@Loki123

I understand that. Why does a function need a critical point to change? Can't it just change without it?

Differential calculus has NOTHING to do with functions in general. It is limited to the neighborhood of values where a function happens to be differentiable. In other words, calculus is not about all functions so your question starts from a false premise. What you should have asked is why does a differentiable function have to have a critical point where specific kinds of qualitative change in the behavior of that function happen.

 

[Loki123]

The key idea is that a critical point is a point where the sign of the derivative may change. If the derivative is continuous, it can change only by passing through zero (though it doesn't have to change when it is zero). If the derivative is undefined, anything can happen because it is not continuous, so you need to check whether the sign changes. (That includes the function itself being undefined.)

So the definition of "critical point" is designed precisely to answer your first question above.

what does it mean if a function is continuous?

 

[Dr.Peterson]

what does it mean if a function is continuous?

You know about derivatives, but not about continuity? That's odd.

It means that the function value is equal to the limit, everywhere. Sound familiar? In other words, typically, it means there are no sudden jumps, which would allow it to change sign without being zero in between.

 

[Loki123]

You know about derivatives, but not about continuity? That's odd.

It means that the function value is equal to the limit, everywhere. Sound familiar? In other words, typically, it means there are no sudden jumps, which would allow it to change sign without being zero in between.

Got it. Thanks!

 

[Loki123]

You know about derivatives, but not about continuity? That's odd.

It means that the function value is equal to the limit, everywhere. Sound familiar? In other words, typically, it means there are no sudden jumps, which would allow it to change sign without being zero in between.

I have come across this:
Throughout this section we will assume, unless otherwise noted, that all functions are continuous. However, continuity of a function does not guarantee that its first and second derivatives are continuous.
Can you explain to me what it means if the function is continuous and first and second derivatives are or are not continuous?

 

[Deleted member 4993]

I have come across this:
Throughout this section we will assume, unless otherwise noted, that all functions are continuous. However, continuity of a function does not guarantee that its first and second derivatives are continuous.
Can you explain to me what it means if the function is continuous and first and second derivatives are or are not continuous?

A saw-tooth function is continuous - but its first derivative is not continuous. If you do not know - what a saw-tooth function is - please Google the term.

 

[topsquark]

I have come across this:
Throughout this section we will assume, unless otherwise noted, that all functions are continuous. However, continuity of a function does not guarantee that its first and second derivatives are continuous.
Can you explain to me what it means if the function is continuous and first and second derivatives are or are not continuous?

How about an example? [imath]y = |x|[/imath]. y is continuous but [imath]y' = \dfrac{x}{|x|}[/imath], which has a discontinuity at x = 0.

(The attached graph has |x| in red and x/|x| in dark blue.)

-Dan

 

[JeffM]

I have come across this:
Throughout this section we will assume, unless otherwise noted, that all functions are continuous. However, continuity of a function does not guarantee that its first and second derivatives are continuous.
Can you explain to me what it means if the function is continuous and first and second derivatives are or are not continuous?

It means exactly what it says.

Consider [imath]f(x) = \sqrt{x^2}.[/imath]

How would you prove that it is continuous over (- 1, 1)?

What is [imath]f'(x)[/imath]?

What is [imath]\lim_{h \rightarrow 0^-} \dfrac{f(0 + h) - f(0)}{h}[/imath]?

What is [imath]\lim_{h \rightarrow 0^+} \dfrac{f(0 + h) - f(0)}{h}[/imath]?

So is f'(x) continuous at x = 0?