Question

locate the absolute extrema for f(x)=(x+2)^1/3 on [-4,0] i have the min which i got -1.295 and the max 1.295. but i am not 100 percet sure, help please

Answer 1

if you just need to check your answer, use a graphing calculator. A TI-83/84 is should be handy

Answer 2

well that it what i used, with the -4 and 0, but i dont get the same answer on paper

Answer 3

what is your derivative?

Answer 4

\[1/3(x+2)^-2/3\]

Answer 5

-2/3 is a power

Answer 6

(1/3)(x+2)^(-2/3).

What did you do next?

Answer 7

i plugged in the vaules you got into x which were -4 through 0

Answer 8

??

plugged -4 and 0 in for x of which function? the derivative function or the function itself?

Answer 9

like this 1/3(-4+2)^(-2/3), i plugged in -4 from [-4, 0], for x

Answer 10

you can't do that. Or what you didn't isn't helpful. What you were supposed to do were to find the critical number by setting the derivative equal to 0

Answer 11

what you did*

Answer 12

so then how would i go with that, i am kinda lost right now

Answer 13

did you do f'(x) = 0? You already found f'(x)

Answer 14

yes, but then i confused what would you do to get x alone i mean i know you are suposed to divide it or subtract it, but how with this one

Answer 15

how would you solve (1/3)(x+2)^(-2/3) = 0 ?

Answer 16

yes would you subtract 1/3 from both sides

Answer 17

no

Answer 18

multiply both sides by 3

Answer 19

so then you would get 0=(x+2)^(-2/3)

Answer 20

yes, that is equivalent to 0 = 1/(x+2)^(2/3)

Answer 21

which has no solution. However, -2 would be the critical number since f'(-2) does not exist.

So you have to points to check:
f(-4), f(-2), f(0)

whichever one is the smallest is the abs min
whichever one is the largest is the abs max

Answer 22

so then would you plug in -4 into the x then

Answer 23

yes but not in f'(x) but f(x), the original function

Answer 24

ok thats what i thought

Answer 25

well thank you very much