Stationary points and nature of calculus question, please help! (1 Viewer)

[wootwoot1234]

Find any stationary points on the curve f(x) = x4 − 2x2 − 3. What type of stationary points are they?

I can find the stationary points, but I can't seem to determine their nature/type.

Stationary points are (0,-3), (-1,-4), (1,-4).

If possible, please include explanation/working out.

Any help is greatly appreciated! Thanks in advance

 

[SpiralFlex]

For stationary points,








(Respectively.)

To determine nature you can do it two ways, I will show you the tabular method.

Basically you grab the value of your point and test for the change of sign. By substituting a close value to it from the left and the right, we should usually see a change of sign.

For



is a maximum turning point.


For



is a minimum turning point.


For



is a minimum turning point.


Alternative is the second derivative,






At



is a maximum turning point.


At



is a minimum turning point.


At



is a minimum turning point.

 

[wootwoot1234]

Thank you so much for your help SpiralFlex! You mentioned that there were two methods to determine the nature. Assuming you are talking about the method where you get points on either side of the x-value and substitute it into the original equation, which method is more reliable and can be used for all questions?

Thank you to interesting as well!

 

[interesting]

Hi SpiralFlex, thank you so much for your help! You mentioned that there were two methods to determine the nature. Assuming you are talking about the method where you get points on either side of the x-value and substitute it into the original equation, which method is more reliable and can be used for all questions?

testing either side cos it also tells u if its an inflexion point

 

[SpiralFlex]

Hi SpiralFlex, thank you so much for your help! You mentioned that there were two methods to determine the nature. Assuming you are talking about the method where you get points on either side of the x-value and substitute it into the original equation, which method is more reliable and can be used for all questions?

In this case, the second derivative method would be more "user friendly", you can instantly differentiate and test for the concavity of the graph at those respective points. However things involving fractions with roots are quite tedious to differentiate so we best resort to testing a close value on either side of the curve.

testing either side cos it also tells u if its an inflexion point

Nice first use of Latex.

 

[interesting]

In this case, the second derivative method would be more "user friendly", you can instantly differentiate and test for the concavity of the graph at those respective points. However things involving fractions with roots are quite tedious to differentiate so we best resort to testing a close value on either side of the curve.

how can we tell if it's inflexion points then? when y''(x)=0 AND y'(x)=0 ?

 

[wootwoot1234]

testing either side cos it also tells u if its an inflexion point

Thanks so much! You have both been a great help.

How do you rep someone?

 

[Alkanes]

If there is a change in concavity. You can use the table of values.

 

[SpiralFlex]

how can we tell if it's inflexion points then? when y''(x)=0 AND y'(x)=0 ?

We have to resort back to the tabular method then (if it's horizontal inflexion). You can choose which method you use unless otherwise stated. But with thee equation, I can tell they will not give inflexions by inspecting the curve. If it's inflexion then we test for the change in concavity.

 

[wootwoot1234]

If there is a change in concavity. You can use the table of values.

By looking at the curve, how can you tell if there is a change is concavity?

 

[SpiralFlex]

By looking at the curve, how can you tell if there is a change is concavity?

We use the second derivative tabular method. If there is an possible oblique inflexion. (We just call this an inflexion.)

This is possible when resorting back to our table, we have to test it using the second derivative to see if it changes sign, if it does, it is an inflexion. If it does not, we need to resort back to our first table as our second derivative test has fail, we must test the nature of it manually.

 

[wootwoot1234]

We have to resort back to the tabular method then (if it's horizontal inflexion). You can choose which method you use unless otherwise stated. But with thee equation, I can tell they will not give inflexions by inspecting the curve. If it's inflexion then we test for the change in concavity.

Can you tell it isn't an inflexion by inspecting the curve before it is differentiated and its nature is determined?

 

[interesting]

By looking at the curve, how can you tell if there is a change is concavity?

is negative concavity
is positive concavity
when there's a change in concavity at a point it would have a combination of both kind of like the curve

 

[wootwoot1234]

We use the second derivative tabular method. If there is an possible oblique inflexion. (We just call this an inflexion.)

This is possible when resorting back to our table, we have to test it using the second derivative to see if it changes sign, if it does, it is an inflexion. If it does not, we need to resort back to our first table as our second derivative test has fail, we must test the nature of it manually.

Oh ok, so both methods may have to be used to answer the question?

 

[SpiralFlex]

Can you tell it isn't an inflexion by inspecting the curve before it is differentiated and its nature is determined?

By sketching the polynomial roughly...I guess. But you should not do that in examinations, they expect you to use these methods.

Oh ok, so both methods may have to be used to answer the question?

First method always answers the question except for (inflexions - oblique). Second derivative may answer the question. Do a few more practice questions and you'll see what I mean. If you have the Cambridge book, I recommend doing the exercise on stationary points.

 

[interesting]

By sketching the polynomial roughly...I guess. But you should not do that in examinations, they expect you to use these methods.



First method always answers the question except for (inflexions - oblique). Second derivative may answer the question. Do a few more practice questions and you'll see what I mean. If you have the Cambridge book, I recommend doing the exercise on stationary points.

umm the first method does work for inflexions?
if you find that both sides have positive dy/dx or negative dy/dx

 

[SpiralFlex]

umm the first method does work for inflexions?
if you find that both sides have positive dy/dx or negative dy/dx

It does not work for oblique inflexions but horizontal ones. If we strictly define inflexions as oblique only. Normally we just say inflexions.

 

[wootwoot1234]

By sketching the polynomial roughly...I guess. But you should not do that in examinations, they expect you to use these methods.



First method always answers the question except for (inflexions - oblique). Second derivative may answer the question. Do a few more practice questions and you'll see what I mean. If you have the Cambridge book, I recommend doing the exercise on stationary points.

Oh that brings me to another question. Which textbook would you recommend, Cambridge or Fitzpatrick for Mathematics 2 unit?

I just started year 12 Calculus, so i'm still a little fuzzy, as they say, about some concepts.

Thanks again guys!

 

[Alkanes]

Fitzpatrick for 2U is good

 

[interesting]

Oh that brings me to another question. Which textbook would you recommend, Cambridge or Fitzpatrick for Mathematics 2 unit?

I just started year 12 Calculus, so i'm still a little fuzzy, as they say, about some concepts.

i like fitz