Question

hi i really need help with this calc question.
find the extrema, points of inflection, intervals where incre/dec, intervals where concave up or down, and any vertical/horizontal asymp.
y= 2x(x-3)^2

Answer 1

find y' and y'', then the calculus is over

Answer 2

you need to sketch the basic graph from this info?

Answer 3

notice it is a cubic function, it prolly turns around a couple times inc/dec

Answer 4

What you get for the first derivative? product and chain rule

Answer 5

okay this is what i got

Answer 6

as my first direvitive= 2(c-30^2 +2(x-3)x

Answer 7

@DanJS also i need help with the second derivative.

Answer 8

man, sorry, i dont get notifications

Answer 9

yes, me too

Answer 10

okay so if my first part is correct which i do no tthink it is. plz tell me how to do the derivative

Answer 11

the function is a product of two terms, 2x and (x - 3)^2, recall the power rule

[f*g] ' = first * derivative second + second * derivative first

Answer 12

and the chain rule for the (x-3)^2 term when you have to take it's derivative

Answer 13

2(x-3)? so my first derivative is right?

Answer 14

\[y' = (2x)*2(x-3)^1*1 + (x-3)^2 * 2\]
\[y' = 6*(x-3)*(x-1)\]

Answer 15

and critical points there would be 3 and 1

Answer 16

cleans up good after expanding all that out and simplify

Answer 17

yes critical points to test where maybe the thing changes from increaing to decreasing, are where the slope is zero there (horizontal),

y ' = 0, yep, when xis 3 or 1

Answer 18

so what about the second derivitive?

Answer 19

x = 3, point y = 0
x = 1, y = 8

(3,0) and (1,8) you can add to the graph, possible change in directions

Answer 20

just pick any easy number to use inside each of those intervals between the critical points, and see if the derivative (slope) is positive or negative

Answer 21

yes, that how i am doing it too. thank that so nice of you to help me. :) tysm. but i have a question.

Answer 22

(2x)∗2(x−3)1∗1+(x−3)2∗2
how did u got 6(x-3)(x-1) from the equation above
y′=6∗(x−3)∗(x−1)

Answer 23

are u there?

Answer 24

sorry had to do somthing back

Answer 25

expand it all out, and combine like terms, then factor, i think , (i just used calculator)

Answer 26

it's okay. i wiill do anything to get done with this problem today lol. okay so this is what i had 2((x-3)^2 +2(x+3)x)

Answer 27

how did you get that?

Answer 28

wait let me attach a file

Answer 29

that equals 6(x^2+3),

Answer 30

are u able to open it?

Answer 31

[2x] ' = 2

[(x-3)^2] ' = 2(x-3)^1 * 1

Answer 32

did u open the link?

Answer 33

oh are you doin the second derivative

Answer 34

no, i am talking about the first

Answer 35

did u see if the first derivitive u got is right or not with the one i got from the website?

Answer 36

let me look at the site, pretty sure mine is right, it matches the graph

Answer 37

yeah, i think u are right too but just making sure.

Answer 38

yeah that last answer is the same, just messy looking

Answer 39

so yours is right

Answer 40

yeah, so is the website in an uglier form, you just copied it here wrong, so it wasnt good..

Answer 41

but it's okay if i write yours?

Answer 42

yeah i would, that is probably what they want in your class or book

Answer 43

okay. thanks! i still need help with the second derivitive

Answer 44

y = 2x(x-3)^2

y' = 6*(x-3)*(x-1)

y'' = 12x - 24

Answer 45

using the chain rule, pretty simple, the derivative of those quantities is 1, and th equantity to the zero power is 1

or you can multiply it all out, and just use the power rule

Answer 46

ao y "= 1

Answer 47

so*

Answer 48

y'' = 6 * [ (x-3) * 1*(x-1)^0 *1 + (x-1)* 1 *(x-3)^0 *1 ]

Answer 49

that is the chain rule, more typing than thinking ,

6*[(x-3) + (x-1)]

12x - 24

Answer 50

If you expand y' first, you get

y ' = 6x^2 - 24x + 18

y ' ' = 12x - 24 (easy power rule in that case)

Answer 51

thanks but the answer whould be 12(x-2) critical point at 2

Answer 52

lol yeah, factor a 12 out

Answer 53

for the y"