Stationary Points (3 Viewers)

[Smile12345]

Aah I havn't done this topic and I'm afraid of giving you incomplete or wrong help... but I'll try:

The first derivative describes the gradient function of the original function (i.e. how it grows and decays). The second derivative describes the gradient function of the first derivative. So how does the second derivative describe the original function? Well that's quite hard to see if you're not that great at visualising but basically you map the point from the original to the first derivative to the second (best to generally memorise here imo).

A point of inflexion is where the graph changes concavity. Concavity basically describes the rate of change of the gradient.
If you picture y = x^2: It's concavity is upwards and hence its gradient always increases (it starts from very negative and then exponentially becomes very positive).

We can say a point of inflexion is a max/min point of the first derivative. This is harder to understand but if you think about it maybe you will see why its so... (sorry this is the limitations of my expression - I'm not a great teacher).

No worries... Thanks.

 

[Smile12345]

If you have a point of inflection, y''=0.
So differentiate to get your f'''(x), then sub in that value of x.
If it's a point of inflection at (1,-2) then f''(1)=0 and f(1)=-2.

Then simultaneous with those two equations.

Yes, this is what I was looking for... Thanks.

 

[Smile12345]

AwkwardAnchor is right, but what I was trying to get at is that the values of x for which y"=0 are not necessarily the points of inflexion on y. I believe understanding the derivatives and its implications to the original function is far more vital than arriving to the answer.
Good luck, goodnight

Yes, I understand where you are coming from.... It will be a point of infexion but it may not be a point of horizontal inflexion...
Thanks... Good Afternoon.

 

[Smile12345]

Would you like some more help? I can explain the 'reasoning' or whatever behind it, if you wish.

Are you referring to me???

 

[seanieg89]

It will be a point of infexion but it may not be a point of horizontal inflexion...

Not quite, for example, the function y=x^4 has y''(0)=0, but 0 is NOT a point of inflexion of y=x^4, as y=x^4 just looks like a parabola.

The key is that concavity must CHANGE on either side of a point of inflexion. Ie, we are looking for maxima and minima of y' rather than just looking for stationary points of y'.

 

[Smile12345]

Not quite, for example, the function y=x^4 has y''(0)=0, but 0 is NOT a point of inflexion of y=x^4, as y=x^4 just looks like a parabola.

The key is that concavity must CHANGE on either side of a point of inflexion. Ie, we are looking for maxima and minima of y' rather than just looking for stationary points of y.

Yeah ok then.... Thanks for that...

I was thinking about a question like this... (I'll put up how I have written to do it)

'a) Show that . has a point of inflexion at x = 1.'
'b) Is it a horizontal point of inflexion'

 

[Smile12345]

a)


Let y'' = 0
6x - 6 = 0
6x = 6
x = 1

Check if concavity changes:
x = 0 1 2
y'' = -6 ? 6

Therefore concavity changes so there is a point of inflexion at x = 1

b) When x = 1, y' = 3(1)^2 - 6(1) -6
= -9 (doesn't = 0)

Therefore x = 1 is not a horizontal point of inflexion

Or is this wrong?

 

[seanieg89]

a)


Let y'' = 0
6x - 6 = 0
6x = 6
x = 1

Check if concavity changes:
x = 0 1 2
y'' = -6 ? 6

Therefore concavity changes so there is a point of inflexion at x = 1

b) When x = 1, y' = 3(1)^2 - 6(1) -6
= -9 (doesn't = 0)

Therefore x = 1 is not a horizontal point of inflexion

Or is this wrong?

Yep sure.

 

[Smile12345]

Yep sure.

Ok. So what I've written is correct?

 

[anomalousdecay]

a)

Check if concavity changes:
x = 0 1 2
y'' = -6 ? 6

The question mark is zero.

 

[Smile12345]

The question mark is zero.

Please explain... Sorry.

 

[nerdasdasd]

Please explain... Sorry.

Y" = 6x - 6

Sub x = 1...

 

[anomalousdecay]

Please explain... Sorry.

Substitute x=1 into y''. You will get:

 

[Smile12345]

Substitute x=1 into y''. You will get:

Yes, I get it now..., Is what I said right?

 

[anomalousdecay]

Yes, I get it now..., Is what I said right?

Yes. Your method and tests were in the right order and are all correct. Well done.

 

[Smile12345]

Yes. Your method and tests were in the right order and are all correct. Well done.

Ok... Awesome... Thanks... I just left it as a ? because I didn't necessarily need the value...