Question

the max value of
f(x)= 2x^3-9x^2+12x-1 on [-1,2] is

Answer 1

any ideas on how to begin?

Answer 2

mm, take the first deriv and set equal to 0?
i know that's what you do to find max value... but, idk i can't seem to get the answer correct

Answer 3

ok that's a good start

Answer 4

setting = 0 will tell you the critical points

Answer 5

did you remember to check the end pts?

Answer 6

which are the end points? >.< lol

Answer 7

\[f'\left( x \right)=6x^2-18x+12=0~gives~x^2-3x+2=0,x^2-2x-x+2=0\]
\[x \left( x-2 \right)-1\left( x-2 \right)=0,\left( x-2 \right)\left( x-1 \right)=0,x=1,2\]
check at x=-1,1,2

Answer 8

what is your domain?

Answer 9

where is the @surjithayer -1 from?

Answer 10

and how do you check them? :0

Answer 11

jst substitute them back into the equation??

Answer 12

test them, I suggest finding the critical points, then seeeing if the first deriv is neg or pos

Answer 13

you are given the interval[-1,2]
i found two values x=1,x=2

Answer 14

okay, got that!
but how do you test the x values?
it's a non calculator section...

Answer 15

you want to look at the first derivative, what does it represent?

Answer 16

\[f \left( x \right)=2x^3-9x^2+12x-1,find f \left( -1 \right),f \left( 1 \right),f \left( 2 \right)\]

Answer 17

you got that eqn by taking the intergral ..?

Answer 18

@SoccerGirl13 Let's restart, I think you may be getting a little confused. What was your given?

Answer 19

f(x)= 2x^3-9x^2+12x-1

Answer 20

@FibonacciChick666

Answer 21

alright, and what are we evaluating it over?

Answer 22

[-1,2]

Answer 23

good, so the thing to remember is that you always have to check your end points and any critical points on your domain, do you follow?

Answer 24

yea :D I get it (:
it all just clicked :)

Answer 25

thanks (:

Answer 26

np, also, when you can't graph. Check your first derivative, is it positive? that means your graph has a positive slope. Which means increasing, if it is increasing on your domain, you know that the end pt must be the local max