Question

Could someone help verify these answers i ahve: i was asked to find the abso/rel. max and min for this function: y=x^3-3x^2+4 over the interval [-2,4]. I have the rel. max as being 4, the rel. min as 0. Then i have the abs.min as -18 and abso. max as 20.

Answer 1

Am i correct, with what i have

Answer 2

Are you sure what you have is correct?

Answer 3

on your first post x shoudl equal 2 not 4

Answer 4

yes im wrong

Answer 5

y'=3x(x-2) right
lol
i said x=4 not x=4
that was an oops
x=0
x=2

Answer 6

x=0, y=4
x=2, y=2^3-3(2^2)+4=8-12+4=0

Answer 7

y''=6x-6
y''(0)=-6 => max
y''(2)=6(2)-6=12-6=6 => min

Answer 8

you can also use the first deriavtive
y'=3x(x-2) and use test points around the critical numbers intead of using 2nd derivative

Answer 9

to find max and mins

Answer 10

Rel. max: 4 at x=0, Rel. min: 0 at x=2

Abso max: 20 at x=4, Abso min: -18 at -2

Is this correct?

Answer 11

f'(-1)=+ f'(1)=- f'(3)=+
---------|------|----
0 2
^ U
so max at x=0
and min at x=2

Answer 12

now is that the relative max and min or abso max and min

Answer 13

or is it both, since 0 and 2 occur within the invtervals

Answer 14

oh we are looking at an interval

Answer 15

x=0, y=4
x=2, y=0
x=-2, y=-16 abs min at x=-2
x=4, y=20 abs max at x=4

Answer 16

so is there no relative min and max at 2 and 0

Answer 17

there is
i already said they were above
did you already forget lol?

Answer 18

i said rel min at x=2
and rel max at x=0

Answer 19

no, i didnt forget, i just want ot make sure that there is an indeed a rel max and min, and abso max and min,

Answer 20

your funny

Answer 21

no u lol

Answer 22

so do you have any questions?

Answer 23

i didn't read your whole paragraph above at first so i didn't know we were just looking at an interval

Answer 24

find max and min of y=(x-1)^2(x+1)^2 smarty pants