Question

Find the points of inflection: y=x^3-8x^3+x

Answer 1

find the second derivative of the function and set it equal to zero, this will give you the x value for the points of inflection

Answer 2

So far i got the derivative, 4x^3 -24x^2 + 1

Answer 3

ah yes sorry about that.. ^^"

Answer 4

find the 1st and 2nd derivatives

solve the 1st derivative to find the stationary points

find the 2nd derivative...

test the stationary points

f''(a) > 0 min

f"(a) = 0 horizontal pt of inflexion

f"(a) < 0 a maximum...

if there isn't a horizontal point of inflexion then solve the 2nd derivative.

then test there is a change in concavity either side of the solution.

if there is then you have a point of inflexion.

Answer 5

the equation is x^4-8x^3 + x

Answer 6

@campbell_st oh so to find the Point of Inflection or (P.I.) I would have to find the 2nd Derivative?

Answer 7

I thought to find P.I. i would only have to do First Derivative.. :o

Answer 8

yes... but you need to find the solutions to the 1st derivative as your point of inflexion may be a horizontal point...

so the way I do it is

find 1st derive

then 2nd derive

solve 1st derive = 0

test the solutions in the 2nd derive by substituting.

then solve the 2nd derive...

test for a change in concavity by picking a value close to the solution on either side... and seeing a change in the sign.

Answer 9

there are 2 types of inflexion points....

1. horizontal... found at the 1st derive ann tested in the 2nd derive.

2. just point of inflexion... found as a solution to the 2nd derive and then test.

Answer 10

hope it make sense

Answer 11

ah yes iremember now! :D

Answer 12

yeah i was just too lazy to look over my notes lol xD

Answer 13

and just graphing your curve there is a horizontal pt of inflexion and a normal one...

.

Answer 14

@campbell_st can you help me with another one? ^^"

Answer 15

ok...

f'(x) = 4x^3 - 24x^2 + 1

oops may have made a mistake... no obvious solutions there...

f"(x) = 12x^2 - 48x

solutions to f'(x) are x = 0 or 4

check change in concavity

x = -1 f''(-1) = 60

x = 1 f''(1) = -36

so pt of inflexion at x = 0

tes x= 4

f''(3) = 108 - 144 = -36

f''(5) = 300 - 240 = 60

change of sign so pt of inflexion at x = 4

Answer 16

yeah i got it now thx :)