The Derivative and Curve Sketching

[eutas1]

Please refer to attachment 'Q1' for the question - I worked out how to do the question (refer to attachment A for my working) but I have some questions.

1. I do not understand the worked solution answer - refer to the red bracket in attachment 'Q1'. Why does a positive gradient function mean no stationary points ???
2. I do not understand the meaning behind it - please refer to attachment Q2 which has my questions.

Thank you!

 

[JeffM]

If f(x) is a differentiable function and f(x) has a stationary point (a local extremum) at a , f PRIME of x at a = 0.

If you had graphed f’(x) properly and at a relevant scale, you would have found 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. It is the value of f’(x) that is relevant. Graph f(x).

 

[eutas1]

If f(x) is a differentiable function and f(x) has a stationary point (a local extremum) at a , f PRIME of x at a = 0.

If you had graphed f’(x) properly and at a relevant scale, you would have found 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. It is the value of f’(x) that is relevant. Graph f(x).

But how can a turning point (or inflection point) NOT have a gradient of 0? I thought all turning/inflection points had a gradient of 0 and therefore are all stationary points?????

 

[pka]

But how can a turning point (or inflection point) NOT have a gradient of 0? I thought all turning/inflection points had a gradient of 0 and therefore are all stationary points?????

Your difficulty is in understanding the terms being used. Points at which a differentiable function has derivative zero are extrema or inflection points.
But inflection points are not an extrema point. There are two types of inflection points: a rising point and a falling point. But nevertheless the derivative is zero. SEE HERE.

 

[JeffM]

But how can a turning point (or inflection point) NOT have a gradient of 0? I thought all turning/inflection points had a gradient of 0 and therefore are all stationary points?????

[MATH]f(x) = x^3 + 4x - 1 \implies f’(x) = 3x^2 + 4 \implies f’’(x) = 6x \implies f’’(0) = 0.[/MATH]
If f(x) has a stationary point (a turning point, a local extremum), then at that point its first derivative must have a VALUE equal to zero.

IT IS FALSE THAT

[MATH]a \text { is a stationary point of the first derivative of } f(x) \implies a \text { is also a stationary point of } f(x).[/MATH]
It is obvious that [MATH]3x^2 + 4 \ge 4 > 0.[/MATH]
There is consequently no stationary point of f(x) because there is no point where its first derivative is zero. It is true that f’(x), the derivative of f(x), has a stationary point, and that the first derivative’s derivative, the second derivative, has a value of zero there, but the first derivative does NOT have a value of zero at its stationary piint. This might have benn clearer had you not used such a gigantic scale on your graph.

 

[eutas1]

But inflection points are not an extrema point. There are two types of inflection points: a rising point and a falling point. But nevertheless the derivative is zero. SEE HERE.

How can an inflection point not be a stationary point if the derivative is still zero?

 

[eutas1]

[MATH]f(x) = x^3 + 4x - 1 \implies f’(x) = 3x^2 + 4 \implies f’’(x) = 6x \implies f’’(0) = 0.[/MATH]
If f(x) has a stationary point (a turning point, a local extremum), then at that point its first derivative must have a VALUE equal to zero.

IT IS FALSE THAT

[MATH]a \text { is a stationary point of the first derivative of } f(x) \implies a \text { is also a stationary point of } f(x).[/MATH]
It is obvious that [MATH]3x^2 + 4 \ge 4 > 0.[/MATH]
There is consequently no stationary point of f(x) because there is no point where its first derivative is zero. It is true that f’(x), the derivative of f(x), has a stationary point, and that the first derivative’s derivative, the second derivative, has a value of zero there, but the first derivative does NOT have a value of zero at its stationary piint. This might have benn clearer had you not used such a gigantic scale on your graph.

Hmmm okay... We have not covered second derivatives, but I can see what you are saying about how even if it is a turning/inflection point, if the gradient is NOT 0, then it is NOT a stationary point. I just thought that turning/inflection points ALWAYS have a gradient of 0 [f'(x) = 0]
I guess this is something I just have to remember..
Thank you!

 

[pka]

How can an inflection point not be a stationary point if the derivative is still zero?

The answer to your question is: it depends who you ask.
The descriptive "stationary" has some unfortunate meanings. It means different things in different situations. Again SEE HERE.

 

[eutas1]

The answer to your question is: it depends who you ask.
The descriptive "stationary" has some unfortunate meanings. It means different things in different situations. Again SEE HERE.

"a stationary point of a differentiable function of one variable is a point on the graph of the function where the function's derivative is zero"
^ If you use this definition, then wouldn't any turning/inflection point with a derivative of zero be a stationary point???

 

[nasi112]

This is the graph of the function [MATH]f(x) = x^3 + 4x^2 - 1[/MATH]


It is obvious that it has no maximum or minimum points. It has an inflection point at [MATH](0, -1)[/MATH]
Do you see any zero slope at that point? Absolutely no. In fact, inflection point is where the graph will change from concaving up to concaving down or the reverse. Also, we did not find the inflection point when we set the first derivative to zero, we found it when we set the second derivative to zero. First derivative is dealing with stationary points, not second derivative.

If you have always thought that inflection points have a gradient of zero, this means you have never looked at graphs, or you have never understood graphs.

I will give you also another important information. Finding stationary points and inflection points is the main key to draw graphs precisely without using any technology. Of course drawing things in hand will not be as good as technology, but if you leave scaling behind, you can graph any function and understand its properties using stationary points and inflection points.

 

[pka]

"a stationary point of a differentiable function of one variable is a point on the graph of the function where the function's derivative is zero" ^ If you use this definition, then wouldn't any turning/inflection point with a derivative of zero be a stationary point???

Did you read the link. ][3] A stationary is Informally, is a point where the function "stops" increasing or decreasing (hence the name).
At a point of inflection the slope does not stop, it just changes.

 

[eutas1]

This is the graph of the function [MATH]f(x) = x^3 + 4x^2 - 1[/MATH]
View attachment 26729

It is obvious that it has no maximum or minimum points. It has an inflection point at [MATH](0, -1)[/MATH]
Do you see any zero slope at that point? Absolutely no. In fact, inflection point is where the graph will change from concaving up to concaving down or the reverse. Also, we did not find the inflection point when we set the first derivative to zero, we found it when we set the second derivative to zero. First derivative is dealing with stationary points, not second derivative.

If you have always thought that inflection points have a gradient of zero, this means you have never looked at graphs, or you have never understood graphs.

I will give you also another important information. Finding stationary points and inflection points is the main key to draw graphs precisely without using any technology. Of course drawing things in hand will not be as good as technology, but if you leave scaling behind, you can graph any function and understand its properties using stationary points and inflection points.

How is that the graph of f(x) = x^3 +4x^2 -1 ?? f(x) is meant to be a cubic function, not a linear one?

 

[eutas1]

Did you read the link. ][3] A stationary is Informally, is a point where the function "stops" increasing or decreasing (hence the name).
At a point of inflection the slope does not stop, it just changes.

I thought the slope DOES stop at at a point of inflection....

 

[nasi112]

How is that the graph of f(x) = x^3 +4x^2 -1 ?? f(x) is meant to be a cubic function, not a linear one?

If the point [MATH](0,-1)[/MATH] is an inflection point, you should understand this is not a linear function.

It is not so obvious it is not linear because it was zoomed in. I zoomed it in just to show you where the inflection point is

Here is another graph for the same function zoomed out.

 

[eutas1]

If the point [MATH](0,-1)[/MATH] is an inflection point, you should understand this is not a linear function.

It is not so obvious it is not linear because it was zoomed in. I zoomed it in just to show you where the inflection point is

Here is another graph for the same function zoomed out.

View attachment 26730

Wait but if you zoom in, it looks like there are max/min turning points?

 

[nasi112]

Wait but if you zoom in, it looks like there are max/min turning points?

View attachment 26731

lol, by mistake i have written [MATH]f(x) = x^3 + 4x^2 - 1[/MATH]
It should be [MATH]f(x) = x^3 + 4x - 1[/MATH]
The graph I have sent to you is for [MATH]f(x) = x^3 + 4x - 1[/MATH]. Try and let me know.

 

[JeffM]

For gracious sake.

If f(x) is a “stationary point” at x = a, f’(a) = 0 and a is a turning point of f(x).

It simply is not true that there is a “turning point” of f(x) where there is a “turning point” of f’(x).

I think pka is correct that you are using terms loosely and thereby confusing yourself.

What do you mean EXACTLY by “turning point”? What do you mean EXACTLY by “stationary point”? Frankly, I would not use the phrase “stationary point” outside the context of dynamics, but your book seems to do so.

Using the phraseology of your book, we can say

[MATH]\text {If a differentiable function } f(x) \text { has a stationary point at a, then its first derivative at a necessarily} = 0.[/MATH]
Now is there any doubt in your mind that the first derivative of this function is [MATH]3x^2 + 4?[/MATH]
Is there any value of x where that derivative could possibly equal zero?

If the answer is to that is no, how could f(x) have a stationary point?

You say that the gradient of the gradient is zero at x = 0. True, but irrelevant. We are not interested in the gradient of the gradient but rather the value of the gradient. Moreover, as I say now for the third time, the scale on which you have graphed, not the function, but its gradient, is at too gross a scale to show that the gradient is always positive. Do you pay any attention to what is said to you?

 

[eutas1]

For gracious sake.

If f(x) is a “stationary point” at x = a, f’(a) = 0 and a is a turning point of f(x).

It simply is not true that there is a “turning point” of f(x) where there is a “turning point” of f’(x).

I think pka is correct that you are using terms loosely and thereby confusing yourself.

What do you mean EXACTLY by “turning point”? What do you mean EXACTLY by “stationary point”? Frankly, I would not use the phrase “stationary point” outside the context of dynamics, but your book seems to do so.

Using the phraseology of your book, we can say

[MATH]\text {If a differentiable function } f(x) \text { has a stationary point at a, then its first derivative at a necessarily} = 0.[/MATH]
Now is there any doubt in your mind that the first derivative of this function is [MATH]3x^2 + 4?[/MATH]
Is there any value of x where that derivative could possibly equal zero?

If the answer is to that is no, how could f(x) have a stationary point?

You say that the gradient of the gradient is zero at x = 0. True, but irrelevant. We are not interested in the gradient of the gradient but rather the value of the gradient. Moreover, as I say now for the third time, the scale on which you have graphed, not the function, but its gradient, is at too gross a scale to show that the gradient is always positive. Do you pay any attention to what is said to you?

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.

 

[nasi112]

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.

Turning points always have a gradient of zero. It is necessary that turning points have a gradient of zero. Inflection points are not turning points. This is the first thing that you have to understand.

Inflection points are always not stationary points because they are not turning points. This is the second thing you have to understand.

And yes, stationary points have always the first derivative equal to zero. FIRST DERIVATIVE. Not second, not third, not fourth. Only the first derivative. This is the third thing you have to understand.

Reading your paragraph proves that you are still confused.

If you cannot understand these three points NOW, just memorizing them. One day, you will understand them. They are not difficult.

 

[eutas1]

Turning points always have a gradient of zero. It is necessary that turning points have a gradient of zero. Inflection points are not turning points. This is the first thing that you have to understand.

Inflection points are always not stationary points because they are not turning points. This is the second thing you have to understand.

And yes, stationary points have always the first derivative equal to zero. FIRST DERIVATIVE. Not second, not third, not fourth. Only the first derivative. This is the third thing you have to understand.

Reading your paragraph proves that you are still confused.

If you cannot understand these three points NOW, just memorizing them. One day, you will understand them. They are not difficult.

Ohh, okay. Thank you!