Question

Find coordinates of stationary pt.of inflexion on curve given by eqn y=3x^4- 4x^3

Answer 1

dy/dx = 12x^3 - 12x^2

Answer 2

dy/dx=0
12x^3-12^2
12x^(x-1)=0
x=0,1

When x=1, y=-1
Wwhen x=0, y=0
From dy/dx, the points are (0,0) and (1,-1)

Answer 3

d^2y/dx^2= 36x^2 - 24x
36x^2 - 24x = 0
12x(3x-2)=0
x=0,2/3

When x=0, y=0
When x=2/3 , y= -16/27



When x=0, d^2y/dx^2=36(0)^2-24(0) =0
When x=2/3 , d^2y/dx^2 = 36(2/3)^2 - 24(2/3) = 0

The pt.of inflexion are (0,0) and (2/3 , -27/16)

Answer 4

Kinda confuse with my own working.. !

Answer 5

@ikram002p @ganeshie8 @dan815

Answer 6

So, my working is correct ?
@ganeshie8

Answer 7

@ganeshie8

Answer 8

so we have

1)
dy/dx = 0 at the points `(0,0) and (1,-1)`

2)
d^2y/dx^2 = 0 at the points `(0,0) and (2/3 , -27/16)`

Answer 9

lets find out stationary and inflection points now

Answer 10

Okay

Answer 11

How ? Sub value of x into dy/dx and d^2y/dx^2 respectively ?

Answer 12

First of all what is a stationary point ?

Answer 13

It's a point whereyby the gradient of curve is always 0

Answer 14

whereby*

Answer 15

http://mathworld.wolfram.com/StationaryPoint.html

Answer 16

Right, so our stationary points are simply the points at which first derivative is 0 :
` (0,0) and (1,-1)`

Answer 17

Okay

Answer 18

lets find out inflection points

Answer 19

Kk

Answer 20

By using d^2y/dx^=0 rite ?

Answer 21

dx^2 *

Answer 22

yes second derivative is 0 for inflection points
but not all points where second derivative is 0 is an inflection point

Answer 23

its like saying : all dogs have four legs
but not all four legged pets are dogs

Answer 24

we have d^2y/dx^2 = 0 at the points `(0,0) and (2/3 , -27/16)`

that just tells us that inflection occurs only at `few or all` of these points

Answer 25

how do we check whether they are inflection points ?

Answer 26

By testing the sign of d^2y/dx^2 .. : Sub value of x into it

Answer 27

BINGO!

Answer 28

sub a value of x to the left and right of the x value of possible point

Answer 29

to check whether the point `(0,0)` is really an inflection point or not :

1) plugin x = -0.1 into d^2y/dx^2
2) plugin x = +0.1 into d^2y/dx^2

both should give you opposite sign numbers

Answer 30

So, that applies to point (2/3, -16/27)

Answer 31

One is positive the other is negative for both the points.

Answer 32

ince, there's a change of sign, it's a pt of inflexion

Answer 33

Since *

Answer 34

Excellent! that means both are inflection points!

Answer 35

And point (1,1) is a min point coz dy/dx=12 >0 when x=1

Answer 36

Thank you @ganeshie8

Answer 37

Yep! and (0,0) is an inflection point so its neither minimum nor maximum

Answer 38

Okay :)