Question

In this picture

Answer 1

Many choices on how to take the derivative, do you want to multiply it all together to do power rule or chain plus product rule it?

Answer 2

the rules :)

Answer 3

chain sorry lol

Answer 4

Critical numbers are numbers at which f'(a) = 0 or is undefined.
For this equation f(x) =(x+2)(x-1)^2 I'll start by simplifying.
f(x) = (x+2)(x^2-2x+1)
f(x) = x^3 - 2x^2 + x + 2x^2 - 4x + 2
f(x) = x^3 - 3x +2

Answer 5

ok I see what you did on that step

Answer 6

Now it is relatively simplify to take the derivative of,
f'(x) = 3x^2 - 3

Answer 7

Ok

Well we got
\[f(x) = (x-1)^2(x+2)\]

We are doing a product rule that begins with a chain rule so....

\[f'(x) = 2(x-1)^1(1)(x+2) + (1)(x-1)^2\]

Is what we have so far with out resimplifying, follow?

Answer 8

yes i follow :)

Answer 9

Or as jziggy did just multiply it all together haha :P I am sure my way would turn out to be his.

Answer 10

@jziggy where did you get you equation from?

Answer 11

Next, you just need to set the derivative equal to zero, and solve for the critical numbers
and yes as long as we both did the math correctly it would be the same :D

Answer 12

I simplified the equation and used the power rule for each of the terms

Answer 13

HE took the original f(x) and just multiplied it all out and simplified

Answer 14

oh ok

Answer 15

Then used power rule, which is what I would do since I hate chain and product rule :P

Answer 16

ok cool so after you get to his part you take the derivative and set it to 0

Answer 17

Yep that gets you your critical points, then using those x-values do a 1st Derivative test for upward or downward motion

Answer 18

when you get that you should be able to find your extrema

Answer 19

Usually simplifying first is the best route to take, but back to the question.
Setting the derivative equal to zero gives us
3x^2 - 3 = 0
3x^2 = 3
x^2 = 1
x = 1 or -1
So our critical numbers for this are 1 and -1 these are the points at which the slope of the graph is equal to zero.

Answer 20

ok im starting to fall behind :P i am getting it though

Answer 21

Follow what jziggy just said haha

Answer 22

Only thing still is a first derivative test

Answer 23

Now to find out whether these are relative mins or maxes we use the second derivative of the equation

Answer 24

which is just the derivative of the derivative

Answer 25

^ that works as well lol I was tuaght just do 1st deriv test and if you get a down followed by up thats a min or a up followed by a down thats a max

Answer 26

so, f'(x) = 3x^2 - 3
f''(x) = 6x

Answer 27

ok now do we plug in the crit values to the second derivative?

Answer 28

at x = -1,
f''(-1) = -6 since this is negative that means the graph curves downwards at this point, a.k.a. a relative maximum
at x = 1,
f''(1) = 6 since it's positive it means it curves upwards and is a relative minimum.

Answer 29

Well said jziggy well said :P

Answer 30

awesome thanks for the help guys :) i have 7 more to do if u wanna help then I can help you some of your work if you need

Answer 31

I can help with some more if you need, but I've already taken my finals for the year :D

Answer 32

I will be more then happy to help if needed

Answer 33

For me we just finished Derivatives and went to Integration so still frehs in my mind

Answer 34

awesome thank you sooooo much guys i am stressing because i have to be done with this unit and another one by tomorrow night

Answer 35

Calculus can be stressful but when you get it, it can become as easy as Algebra

Answer 36

yeah

Answer 37

Yeah, but make sure to learn the concepts and not just memorize equations ;p

Answer 38

ok im opening the next question

Answer 39

alright

Answer 40

feel free to tag us in it!