differentiation turning points help

x=2 or x=?

Second differential (or alternate methods ... depends on which you have been taught)

Reply 2

Original post by dongonaeatu

3x^2 -6x =0 \implies x^2 - 2x=0 \implies x(x-2)=0

You should get two

x

values.

Reply 3

Fing4

Second derivative test?

Reply 4

Original post by TenOfThem

x=2 or x=?

Second differential (or alternate methods ... depends on which you have been taught)

Original post by raheem94

3x^2 -6x =0 \implies x^2 - 2x=0 \implies x(x-2)=0

You should get two

x

values.

x(x-2) so x=2 or x=0

so d2y/dx2= 6x-6

so when x=0 d2y/dx2= -6 so its a maximum?

when x=2 d2y/dx2= 6(2)-6 = 6 so x=2 is a minmum

is this right

Reply 5

Original post by dongonaeatu

is this right

What do you think?

Perhaps a quick sketch of the graph will confirm (or not)

Reply 6

Original post by dongonaeatu

x(x-2) so x=2 or x=0

so d2y/dx2= 6x-6

so when x=0 d2y/dx2= -6 so its a maximum?

when x=2 d2y/dx2= 6(2)-6 = 6 so x=2 is a minmum

is this right

See the graph of the function:

If you still can't ensure that you are right or wrong, then see the spoiler.

Spoiler

Reply 7

Original post by dongonaeatu

x(x-2) so x=2 or x=0

so d2y/dx2= 6x-6

so when x=0 d2y/dx2= -6 so its a maximum?

when x=2 d2y/dx2= 6(2)-6 = 6 so x=2 is a minmum

is this right

Also remember that you need to find the y-coordinates of the stationary points as well.

Reply 8

Original post by TenOfThem

What do you think?

Perhaps a quick sketch of the graph will confirm (or not)

Original post by raheem94

See the graph of the function:

If you still can't ensure that you are right or wrong, then see the spoiler.

Spoiler

yay! thank you guys

Reply 9

Original post by raheem94

Also remember that you need to find the y-coordinates of the stationary points as well.

ohh ok so i just substitute x into y. so

when x=0 y=1 so (0,1)
when x=2 y= (2)^3-3(2)^2+1 y=-3 so (2,-3)

Reply 10

Original post by dongonaeatu

ohh ok so i just substitute x into y. so

when x=0 y=1 so (0,1)
when x=2 y= (2)^3-3(2)^2+1 y=-3 so (2,-3)

Reply 11

part ii (8 marks)
Show that the tangent to the curve at the point where x=-1 has gradient 9.
Find the coordinates of the other point, P, on the curve at which the tangent has gradient 9 and find the equation of the normal to the curve at P.
Show that the area of the triangle bounded by the normal at P and the x- and y-axes is 8 square units.

so dy/dx is 3x^2-6x
where x=-1 dy/dx= 3(-1)^2-6(-1)=9 so i have shown the tangent to the curve at x=-1 has gradient 9.

now what do i do

Reply 12

Original post by raheem94

please look up at part ii

Reply 13

Find the equation of the tangent

Note ... you know a point and the gradient

Reply 14

Original post by TenOfThem

Find the equation of the tangent

Note ... you know a point and the gradient

i need to find the coordinates of P on the curve at which the tangent has gradient 9

so is it 3x^2-6x=9

so 3x^2-6x-9=0 then i factorise?

Reply 15

Original post by dongonaeatu

please look up at part ii

To find the other point, solve,

3x^2-6x=9

Also find the y-coordinate.

Then write the gradient of the normal, as -1 divided by gradient of tangent.

Then form the equation.

Original post by TenOfThem

Find the equation of the tangent

Note ... you know a point and the gradient

Why find the equation of the tangent, he needs to find the equation of the normal.

Reply 16

Original post by dongonaeatu

i need to find the coordinates of P on the curve at which the tangent has gradient 9

so is it 3x^2-6x=9

so 3x^2-6x-9=0 then i factorise?

Its obvious that you need to factorise it and find the solutions.

By the way, have you read the complete chapter?
Or do you want us to help you on every step?

Reply 17

Original post by raheem94

To find the other point, solve,

3x^2-6x=9

Also find the y-coordinate.

Then write the gradient of the normal, as -1 divided by gradient of tangent.

Then form the equation.

Why find the equation of the tangent, he needs to find the equation of the normal.

I was just going a different way

:smile:

Reply 18

Original post by raheem94

Its obvious that you need to factorise it and find the solutions.

By the way, have you read the complete chapter?
Or do you want us to help you on every step?

Original post by TenOfThem

I was just going a different way

:smile:

sorry, i just find you guys are very good at explaining and it helps me

so 3x^2-6x-9=0 (3x+3)(x-3) so x=3 put this into y so y=1 so p is (3,1)

equation of normal to P so gradient of normal= -1/9

y-1=-1/9(x-3)
9y-9=-x+3
equation of normal to the curve at p is 9y=-x+12
how do i show that the area of the triangle bounded by the normal at P and the x axis is 8 square units

Reply 19