Question

Find all of the critical numbers of the function fx=x^3-3x^2 on the open interval (-1,1).

(Hint: If there is more than one value, separate the critical numbers with commas.)

Answer =

Answer 1

what our derivative?

Answer 2

hold on

Answer 3

3x^2-6x

Answer 4

do we want criticals for mina nd max, or for inflections as well?

Answer 5

just the critical for min and max

Answer 6

then 3x^2-6x=0 will be a good critical points to check for min and max

Answer 7

3x(x-2) = 0 when x=0 and x=2 i beleive

Answer 8

x=2 and x=0 but that is not the correct answer idk why

Answer 9

then our definition of critical points is off

Answer 10

its on the interval (-1,1)

Answer 11

then 0 would be a point of interest

Answer 12

whats our second derivative?

Answer 13

6x-6

Answer 14

http://www.wolframalpha.com/input/?i=x%5E3-3x%5E2+from+x%3D-1+to+1

our graph is this

then x=1 looks to be a point of inflection as well

Answer 15

since we have open end points we cant really include them can we?

Answer 16

yes i guess

Answer 17

but there is def. an answer to this

Answer 18

then it appears out critical points is when x=0 ....

Answer 19

all the others are out of our range of interval

Answer 20

i still cant figure out the answer

Answer 21

f(0) = ?

Answer 22

0

Answer 23

i assume this is a program that your inputing an answer into ...

Answer 24

correct

Answer 25

then it might be a good idea to capture a screen shot with the instructions so that we might be able to formulate a "proper" response :)

Answer 26

it works thanks

Answer 27

lol :) youre welcome