Question

Calc Help! find the points of inflection and discuss the concavity of of the graph of the function......f(x)=x(x-4)^3

Answer 1

Let's review real quick.
If we set the FUNCTION equal to zero, we get ROOTS.
If we set the 1st DERIVATIVE equal to zero, we find critical points.
If we set the 2nd DERIVATIVE equal to zero, what do we get???? C::::

Answer 2

for the 2nd derivative we get the inflection points? :}}}}}

Answer 3

Yayyy :D And what do inflection points tell us?
THOSE are the points where the function is changing from concave up to concave down, and so on.

Answer 4

Did you find f'(x) yet?

Answer 5

yes.....its (x-4)^2(4x-4)!

Answer 6

f double prime is the one where im having trouble....hey can i ask u a q?

Answer 7

sup :o

Answer 8

just curious r u a guy or girl?

Answer 9

\[f(x)=x(x-4)^3\]\[f'(x)=(x-4)^3+3x(x-4)^2\]
Hmmmmmm this is what I came up with for the derivative, applying the product rule :O
Did you forget the product rule perhaps? We have 2 functions of x here :3

And I'm a boy :P lol

Answer 10

Oh u simplified :3 hah nice!

Answer 11

o. okay :} no i didnt its just that for this particular problem the answer was simplified where (x-3......yup!

Answer 12

really ur a guy? not even!

Answer 13

not even?? XD lol
\[f''(x)=2(x-4)(4x-4)+4(x-4)^2\]
Hmm did you get something like this maybe? :D
I took the derivative of the simplified form that you had found.

Answer 14

hey its cuz w/ all those smiley faces u put i would think but whatever it doesnt matter! :]
i got 2(x-4)((4x-4)+2x-8) i think the tutor that helped me skipped a step??? maybe? i think this is why i was confused:l

Answer 15

buhah XD

Answer 16

:]

Answer 17

Ah yes, like you did to me a sec ago >:O with your simplified form *grumble grumble*

Answer 18

Your tutor was just paying it forward i guess :U

Answer 19

llololol

Answer 20

paying it forward?

Answer 21

blah i dunno :P let's continue lol

Answer 22

k :]