The Derivative and Curve Sketching

[JeffM]

Yes, I understand now that a turning/inflection point does not necessarily mean the gradient is 0 --> I was confused on this because I DID think that all turning/inflection points have a gradient of 0. I see now that is not the case.

I understand that now. What I am confused on is how it is possible for an inflection point to not be a stationary point if the derivative is zero (post #6), because if the derivative is 0, I thought that made it a stationary point.

I am sorry that it seems I do not pay attention to what is said to me, I just struggle with understanding things and oftentimes it takes a lot of explaining for me to finally get it. A lot of the time it is hard for me to understand what I am even reading because obviously my maths ability and understanding is a lot lower than anyone who comments, and sometimes the explanations include things I have not learnt. I am trying my best.

If by “turning point” of a differentiable function, you mean a point where the function has a local extremum, then the first derivative of that function WILL ALWAYS HAVE a value of zero at that point.

If by ”turning poin“ of a differentiable function you mean a point where the function has a local extremum, then a point where the first derivative of that function has a value of zero is EITHER a turning point OF ELSE an inflection point, WHICH ARE COMPLETELY DIFFERENT THINGS.

In this problem, the first derivative of the relevant function NEVER has a value of zero so the relevant function NEVER has a “turning point.”

Your graph was done at such a massive scale that you could not tell from looking at it that the first derivative (gradient) of the relevant function is always greater than zero.

 

[eutas1]

I have re-read all the posts from the beginning and I think I have a more clear understanding, but I want to check just in case I'm wrong:

• By graphing f’(x) at a relevant scale, it can be seen that f’(x) = 3x^2 + 4 does indeed have a minimum (a turning point) when x = 0, but the value of f’(x) at that minimum is 4 rather than zero. Therefore, it is NOT a stationary point because the y-value HAS to be 0 in order for a stationary point to exist for f(x).
  ^ I had not realised until right now, that because f’(x) = 0 defines the existence of a stationary point for f(x), the f’(x) actually represents the Y-VALUE for the gradient function/graph, so the y-value needs to be 0. Very stupid. My apologies. It is now very clear to me that obviously if f’(x) does NOT equal 0 – meaning the y-value is NOT 0 – there is NO stationary point for f(x).
  
  
• Just because there is a turning/inflection point for f’(x), does NOT mean it is a stationary point for f(x); just because there is a stationary point for the gradient function of f(x) does NOT mean it is a stationary point for f(x). In order for this to be true, that point (‘a’) would need to have a y-value of 0 --> that is, f’(a) = 0.
  
  
• ALL turning points (local extrema) of a function have a gradient of 0. Inflection points are not included in this because inflection points are NOT turning points (local extrema). Therefore, if an inflection point exists for f(x), it does NOT automatically mean it has a gradient of 0. But if there is a turning point for f(x), then the gradient is automatically 0.
  ^ BUT, if a point has a gradient of 0 for f’(x), then it COULD be an inflection point.

Is this correct?

 

[nasi112]

Why are you interested in graphing the derivative function, [MATH]f'(x)[/MATH]? Graph the original function, [MATH]f(x)[/MATH]

 

[JeffM]

• By graphing f’(x) at a relevant scale, it can be seen that f’(x) = 3x^2 + 4 does indeed have a minimum (a turning point) when x = 0, but the value of f’(x) at that minimum is 4 rather than zero. Therefore, it is NOT a stationary point because the y-value HAS to be 0 in order for a stationary point to exist for f(x).

• 
  You have it fully and correctly. Moreover, because the f’(x) (the y-value of the derivative) is nowhere negative, the value of f(x) is nowhere decreasing.
  

  ^ I had not realised until right now, that because f’(x) = 0 defines the existence of a stationary point for f(x)

  I am not sure what you mean here, but what you are saying is not correct. If a is a “turning point” (“stationary point,” (a term I dislike), then f’(a) = 0. But if f’(a) = 0, a may not be a turning point; instead a may be an inflection point. I think your failure to be absolutely sure of this will cause you errors.

• the f’(x) actually represents the Y-VALUE for the gradient function/graph, so the y-value needs to be 0. Very stupid. My apologies. It is now very clear to me that obviously if f’(x) does NOT equal 0 – meaning the y-value is NOT 0 – there is NO stationary point for f(x).

  I do not at all like the word “stupid” in this context. There is a certain subtlety in the idea of derivatives. The derivative of a function is another function. It is easy to get confused which function you are talking about. This is one reason I am less than charmed by discussing extremum points, turning points, and stationary points. A consistent vocabulary aids understanding. it is confusing enough talking about a function, its first derivative, its second derivative, and its third derivative, and which is the derivative of another (for example, the third derivative is the first derivative of the second derivative) without multiplying terms to reference a single idea.
  

  Just because there is a turning/inflection point for f’(x), does NOT mean it is a stationary point for f(x); just because there is a stationary point for the gradient function of f(x) does NOT mean it is a stationary point for f(x). In order for this to be true, that point (‘a’) would need to have a y-value of 0 --> that is, f’(a) = 0.

  Having parsed this sentence several times, I think it is TECHNICALLY correct, but quite confusing. “Turning points” are never “inflection points” so the compound “turning/inflection” point is simply an invitation to sloppy thinking. Although there may be some weird exceptions, if the first derivative of f(x) has a y-value of 0 at a, then normally f(y) has either a local extremum or else an inflection at a. It will never have both an extremum and an inflection at the same point.
  

  ALL turning points (local extrema) of a function have a gradient of 0. Inflection points are not included in this because inflection points are NOT turning points (local extrema). Therefore, if an inflection point exists for f(x), it does NOT automatically mean it has a gradient of 0. But if there is a turning point for f(x), then the gradient is automatically 0.
  ^ BUT, if a point has a gradient of 0 for f’(x), then it COULD be an inflection point.

• YES To summarize, if the value of the first derivative at a is zero, the function may have an extremum there or else an inflection there, or else something weird there. The point is that a value of zero in the first derivative at a means something important happens to the function at a. What that something is requires further analysis.

 

[JeffM]

Why are you interested in graphing the derivative function, [MATH]f'(x)[/MATH]? Graph the original function, [MATH]f(x)[/MATH]

I am not sure of the context of the question.

Before graphing calculators, it was common to teach a section on “sketching a curve,” which meant making a reasonable approximation of the curve of a given function. Things to do were to approximate the function’s y-intercept and the zeroes of the function and its first and second derivatives and consider possible asymptotes. On the basis of all that data, you could derive a reasonable idea of the curve. This question seems to be much more limited. It is not clear to me that this OP has even heard of second derivatives or the second derivative test.

It may be that the OP used the graphing capabilities of the calculator to find the zeroes of the first derivative, and then, because of the scale issue, got lost in the weeds of distinguishing between the function and its derivative.

 

[eutas1]

I am not sure what you mean here, but what you are saying is not correct. If a is a “turning point” (“stationary point,” (a term I dislike), then f’(a) = 0. But if f’(a) = 0, a may not be a turning point; instead a may be an inflection point. I think your failure to be absolutely sure of this will cause you errors.

^ ?? Is this not just the same as how if f'(a) = 0, it could be a turning point OR an inflection point?


Having parsed this sentence several times, I think it is TECHNICALLY correct, but quite confusing. “Turning points” are never “inflection points” so the compound “turning/inflection” point is simply an invitation to sloppy thinking. Although there may be some weird exceptions, if the first derivative of f(x) has a y-value of 0 at a, then normally f(y) has either a local extremum or else an inflection at a. It will never have both an extremum and an inflection at the same point.

^ Ah, my bad!!! I should have written it in a more precise way. I meant "turning/inflection point" as in it could be turning point or inflection point, but obviously not both at the same time.

 

[JeffM]

I am not sure what you mean here, but what you are saying is not correct. If a is a “turning point” (“stationary point,” (a term I dislike), then f’(a) = 0. But if f’(a) = 0, a may not be a turning point; instead a may be an inflection point. I think your failure to be absolutely sure of this will cause you errors.

^ ?? Is this not just the same as how if f'(a) = 0, it could be a turning point OR an inflection point?

My response was to your saying that “f’(x) = 0 defines the existence of“ an extremum. As has now been said repeatedly, f’(x) equals zero at a may not indicate that f(a) is an extremum.

You really need to be careful with your words, or you will make mistakes. Assuming that a is in an open interval and f(x) is everywhere differentiable in that same interval, then

If a is an extremum of f(x), f’(a) ALWAYS equals zero.

If f’(x) equals zero at a, f(x) MAY OR MAY NOT HAVE an extremum at a.

It is exactly the same distinction as

If I am eating beef, I am certainly eating meat.

If I am eating meat, I may or may not be eating beef.

You really need to grasp that distinction solidly. And, I repeat, a consistent vocabulary will make it easier to grasp the distinction solidly.

 

[eutas1]

My response was to your saying that “f’(x) = 0 defines the existence of“ an extremum. As has now been said repeatedly, f’(x) equals zero at a may not indicate that f(a) is an extremum.

You really need to be careful with your words, or you will make mistakes. Assuming that a is in an open interval and f(x) is everywhere differentiable in that same interval, then

If a is an extremum of f(x), f’(a) ALWAYS equals zero.

If f’(x) equals zero at a, f(x) MAY OR MAY NOT HAVE an extremum at a.

I meant it defines the existence of a turning point OR inflection point. So:
If f’(x) equals zero at a, f(x) will have a turning point or an inflection point at a.

 

[Deleted member 4993]

I meant it defines the existence of a turning point OR inflection point. So:
If f’(x) equals zero at a, f(x) will have a turning point or an inflection point at a.

Most of the time - but Not always

If f(x) = Constant

f'(x) = 0 for all x's WITHOUT a turning point.

 

[eutas1]

Most of the time - but Not always

If f(x) = Constant

f'(x) = 0 for all x's WITHOUT a turning point.

........ wouldn't x need to be 0 for f'(x) ??

 

[JeffM]

........ wouldn't x need to be 0 for f'(x) ??

No, no, no.

[MATH]c \text { is any real constant, and } f(x) = c \implies[/MATH]
[MATH]f(x + h) = c \implies f(x + h) - f(x) = c - c = 0.[/MATH]
[MATH]\therefore h \ne 0 \implies \dfrac{f(x + h) - f(x)}{h} = \dfrac{0}{h} = 0.[/MATH]
[MATH]\therefore f’(x) = \lim_{h \rightarrow 0} \dfrac{f(x + h) - f(x)}{h} = 0.[/MATH]

 

[HallsofIvy]

I have re-read all the posts from the beginning and I think I have a more clear understanding, but I want to check just in case I'm wrong:

• By graphing f’(x) at a relevant scale, it can be seen that f’(x) = 3x^2 + 4 does indeed have a minimum (a turning point) when x = 0, but the value of f’(x) at that minimum is 4 rather than zero. Therefore, it is NOT a stationary point because the y-value HAS to be 0 in order for a stationary point to exist for f(x).
  ^ I had not realised until right now, that because f’(x) = 0 defines the existence of a stationary point for f(x), the f’(x) actually represents the Y-VALUE for the gradient function/graph, so the y-value needs to be 0. Very stupid. My apologies. It is now very clear to me that obviously if f’(x) does NOT equal 0 – meaning the y-value is NOT 0 – there is NO stationary point for f(x).
  
  
• Just because there is a turning/inflection point for f’(x), does NOT mean it is a stationary point for f(x); just because there is a stationary point for the gradient function of f(x) does NOT mean it is a stationary point for f(x). In order for this to be true, that point (‘a’) would need to have a y-value of 0 --> that is, f’(a) = 0.

• 
  A "stationary" point IS, by definition, a point where the derivative is 0. But a "stationary point" is not necessarily either a turning point or inflection point.
  
  
  

  [*]ALL turning points (local extrema) of a function have a gradient of 0. Inflection points are not included in this because inflection points are NOT turning points (local extrema). Therefore, if an inflection point exists for f(x), it does NOT automatically mean it has a gradient of 0. But if there is a turning point for f(x), then the gradient is automatically 0.
  ^ BUT, if a point has a gradient of 0 for f’(x), then it COULD be an inflection point.

Is this correct?

I think you need to be careful to state that you are talking about a DIFFERENTIABLE function. If a function is not differentiable at a turning point, of course the derivative cannot be 0 there- it does not even exist. An example is f(x)= |x|. x= 0 is a turning point but the derivative is not 0 there because the derivative does not even exist there!