Stationary Points (2 Viewers)

[Smile12345]

Hello All...

Could someone please help me with this q??

'Find any stationary points on the curve y = (4x^2 -1)^4' and determine their nature.'

So the y'= 32x (4x^2 -1)^3

So one stationary point would be when x = 0.... Then....

Thanks in advance.

(Sorry I don't know how to use latex...)

 

[Smile12345]

Then do I put 4x^2 - 1 = 0 and solve for x to get + and - 1/2 ??

 

[QZP]

To determine stationary points: Yes solve 32x(4x^2-1)^3 = 0 (your above comment is right).

To determine their nature: A stationary point is a very general term that can describe: Turning point (i.e. max/min point) or a point horizontal inflexion - or a horizontal line but clearly this isn't the case.

Picture a turning point and a point of horizontal inflexion in your head. What is the difference? How would you test for that difference?

 

[Trebla]

Hello All...

Could someone please help me with this q??

'Find any stationary points on the curve y = (4x^2 -1)^4' and determine their nature.'

So the y'= 32x (4x^2 -1)^3

So one stationary point would be when x = 0.... Then....

Thanks in advance.

(Sorry I don't know how to use latex...)

One way to determine the nature is to find y'' and check the concavity at those points. If y'' < 0 when you sub in the point then it is concave down. Concave down graphs have a maximum turning point. If y'' > 0 when you sub in the point then it is concave up. Concave up graphs have a minimum turning point.

 

[Smile12345]

To determine stationary points: Yes solve 32x(4x^2-1)^3 = 0 (your above comment is right).

Ok good...

 

[Smile12345]

Yup ok then...

 

[HSC2014]

Note that determining whether a turning point is a max/min is not necessarily easiest through the second derivative (due to difficult differentiation) and sometimes you ought to just use a table of values and test the immediate left/right points of the stationary point.

 

[anomalousdecay]

Yup ok then...

See. There are many users out there to help

 

[Smile12345]

Note that determining whether a turning point is a max/min is not necessarily easiest through the second derivative (due to difficult differentiation) and sometimes you ought to just use a table of values and test the immediate left/right points of the stationary point.

Yeah, very true... Thanks for the tip.

 

[Smile12345]

See. There are many users out there to help

Yes! Thanks all.

 

[Smile12345]

How about if I had a question like 'The Curve has a point of inflexion at (1,-2). Find the values of a and b.'

 

[Smile12345]

I would find the first and second derivatives wouldn't I ??

Then would I use simultaneous equations?

Thanks in advance.

 

[HSC2014]

Know your aim Smile12345... Why do you want the first/second derivative? How will that help you?
Believe me, knowing why you do each step in mathematics will take you a long way.

 

[Smile12345]

Know your aim Smile12345... Why do you want the first/second derivative? How will that help you?
Believe me, knowing why you do each step in mathematics will take you a long way.

The first derivative tells us where the curve is increasing, decreasing or stationary and the second derivative tells us where the curve is concave up, concave down and locates points of inflexion

Yes, that's very true... Thanks for your words.

 

[Smile12345]

Know your aim Smile12345... Why do you want the first/second derivative? How will that help you?
Believe me, knowing why you do each step in mathematics will take you a long way.

Can you please give me a hint.

 

[Smile12345]

Please...

 

[HSC2014]

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).

 

[AwkwardAnchor]

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.

 

[HSC2014]

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

 

[AwkwardAnchor]

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