Stationary points and nature of calculus question, please help! (2 Viewers)

[Carrotsticks]

What else could it imply?

I'm guessing you're asking for other examples where the double derivative and first derivative is equal to 0, but not necessarily a HPOI.

The most obvious example is:



We have to consider whether the highest order of the polynomial is odd/even when it comes to analysing HPOI.

For example, the curve...



... has f'(x) and f''(x) = 0 at the origin, but there IS in fact a HPOI there, as opposed to another curve like:



Another example is from a question I posted earlier in the Extension 2 Marathon regarding the curve:



Although as n approaches infinity, the stationary point and point of inflexion converge to the same fixed point, the limiting case is most certainly not a HPOI.

 

[SpiralFlex]

True, how come its not a minimum then?

EDIT: i screwed up guys sorry lol. i learnt something from spiral.

Don't you learn a lot when you see my awesome super saiyan mode (V4)?

 

[Carrotsticks]

One example mainly multiple roots in the form

Where n is an integer greater than or equal to 3. Not necessarily a horizontal point of inflexion will occur when both first and second derivative equals to zero.

A horizontal point of inflexion will occur when n is odd.

 

[SpiralFlex]

A horizontal point of inflexion will occur when n is odd.

Yeah I should have said EVEN integer.

 

[bleakarcher]

God, I love this forum. Love ya Spiral, love ya Carrotsticks. got a lot of respect for you guys.

 

[Annum]

Is it possible to have three stationary points? Im getting three different stationay points, and honestly havnt come across a question that involves three different stationary points. thanks

 

[nightweaver066]

Is it possible to have three stationary points? Im getting three different stationay points, and honestly havnt come across a question that involves three different stationary points. thanks

Yes.

 

[Carrotsticks]

Is it possible to have three stationary points? Im getting three different stationay points, and honestly havnt come across a question that involves three different stationary points. thanks

A polynomial of nth degree can have at most n-1 stationary points.

 

[Annum]

The area of a rectangle is 70m^2. Show that the perimeter of the rectangle is given by p=2x+140/x

Can anyone help me with this please?

 

[SpiralFlex]

I see no pretty diagrams!

 

[Annum]

Thankyou, i feel bad that your always helping me with my math questions

 

[Annum]

when finding stationary points and testing its nature, is it possible to get both as 'horizontal point of inflextion' for both points because thats what im getting for this one
y= x^3/3 - x^2 - 8x + 11
Its telling me to use the first derivative to determine its nature
for the stationary points im getting (4, -15 2/3) and (-2, 20 1/3)

I really hope you can help me thanks.

 

[Carrotsticks]

when finding stationary points and testing its nature, is it possible to get both as 'horizontal point of inflextion' for both points because thats what im getting for this one
y= x^3/3 - x^2 - 8x + 11
Its telling me to use the first derivative to determine its nature
for the stationary points im getting (4, -15 2/3) and (-2, 20 1/3)

I really hope you can help me thanks.

Think about it this way... the cubic you have there has degree 3. Thus, it has '2 tokens'.

One stationary point costs 1 token.

One horizontal point of inflexion costs 2 tokens.

Hence the cubic can only have at most 1 horizontal point of inflexion, or two stationary points.

 

[Timske]

when finding stationary points and testing its nature, is it possible to get both as 'horizontal point of inflextion' for both points because thats what im getting for this one
y= x^3/3 - x^2 - 8x + 11
Its telling me to use the first derivative to determine its nature
for the stationary points im getting (4, -15 2/3) and (-2, 20 1/3)

I really hope you can help me thanks.

you shouldnt get 2 Horizontal point of inflexion

MIN @ (4, -15 2/3) and MAX @ (-2, 20 1/3)

u can sub in x values in f'(x) to the left and right of 4 and - 2 to determine if its increasing or decreasing on both sides

Remember f ' (x) < 0 its decreasing and if f ' (x) is > 0 its increasing

f ' (x) = x^2 - 2x - 8
x values to the left and right of 4 are 3 and 5, sub those in f'(x)
youll find its decreasing at 3 and increasing at 5 so therefore its a min

 

[Annum]

thankyou i understand it now

you shouldnt get 2 Horizontal point of inflexion

MIN @ (4, -15 2/3) and MAX @ (-2, 20 1/3)

u can sub in x values in f'(x) to the left and right of 4 and - 2 to determine if its increasing or decreasing on both sides

Remember f ' (x) < 0 its decreasing and if f ' (x) is > 0 its increasing

f ' (x) = x^2 - 2x - 8
x values to the left and right of 4 are 3 and 5, sub those in f'(x)
youll find its decreasing at 3 and increasing at 5 so therefore its a min