Question

find the points of inflection and discuss the concavity of the graph f(x)=x(x-4)^3

Answer 1

just having issues with derriving

Answer 2

still need help ?

Answer 3

yes pleaseee :)

Answer 4

f(x)=x(x-4)^3
u need to use product rule , tried it ?

Answer 5

yeah. I just realised that ive been doing the product rule but dividing by g(x)^2 (/.\) EMBERASSINggg x)

Answer 6

whats g(x)^2 ??

Answer 7

I combined it with quotient rule. x)

Answer 8

here u only need product rule (fg)' = fg'+f'g
f= x, g = (x-4)^3

Answer 9

im not sure if im deriving right. is the derivative of (x-4)^3,, 3(x-4)^2?

Answer 10

yes

Answer 11

thats correct, go ahead

Answer 12

ok. for f''(x)=3x^2-12x+16
(I factored out a 3)

Answer 13

lets check,
(fg)' = fg'+f'g
f= x, g = (x-4)^3
(x (x-4)^3)' = (x-4)^3 + 3x(x-4)^2
= (x-4)^2 (x-4+3x)
= 4(x-4)^2 (x-1)
did u differentiate this to take 2nd derivative ?

Answer 14

I derived (x-4)^3+3x(x-4)^2 how did u get to the bottom???

Answer 15

i factored out (x-4)^2 , which was there in both terms
(x-4)^3 = (x-4)^2 (x-4)

Answer 16

ohhhhh! Ok

Answer 17

thanks. I can do the rest from here. thanks you!

Answer 18

welcome ^_^