Question

Find critical point(s) f(x)=(x+2)^2(x-1)

Answer 1

I'm not sure what the first derivative will be...

Answer 2

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

?

Answer 3

or

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

Answer 4

The second way

Answer 5

do you know what critical point is?

Answer 6

yes

Answer 7

so, firstly you need to find deretitive.do you know how?

Answer 8

\[f'(x)=2(x+2)(x-1)+(x+2)^{2}\] right or not?

Answer 9

I think you are correct. I wasn't sure if I needed to simplify before deriving it.

Answer 10

Well, i think it doesnt matter when you simplify. NOW you need to simplify, can you do it for me?

Answer 11

working on it...

Answer 12

\[f'(x)=2(x+2)(x-1)+(x+2)^{2}=2(x ^{2}-x+2x-2)+x ^{2}+4x+4=\]

Answer 13

I got \[3x^2+6x+2\]

Answer 14

\[=2x ^{2}-2x+4x-4+x ^{2}+4x+4=3x ^{2}+6x\]

Answer 15

oh, we didnt get the same. maybe I have made a mistake!

Answer 16

oops messed up on the 2

Answer 17

that was my mistake!

Answer 18

oh,okay. so now do you know how to find critical point?

Answer 19

So x=0 and x=-2 is what I got

Answer 20

well,

\[3x ^{2}+6x=0\]
\[x(3x+6)=0\]
x=0 or x=-2


all points change a sign,so both of them are critical