Question

Find the absolute maximum and absolute minimum values of f on the given interval. (Round all answers to two decimal places.)
f(x) = xe^(-x2/128)
[-3, 16]

Answer 1

correction:
f(x) = xe^(-x^2/128(

Answer 2

well kehala first:
use a calculator to find the derivative of this
f(x) = xe^(-x^2/128)

Answer 3

then once u have f', set that equal to 0:
f'(x)=0

Answer 4

and you will get some x values..

Answer 5

then find f'' and plug those x values into f ''

Answer 6

and whenever f '' is negative then u have a max, whenever it is positive u have min

Answer 7

Finally, take all of your x values AND endpoints, which are -3 and 16
and plug all of those into f(x), whenever f(x) is largest -> that's ur abs max and f(x) smallest -> abs min

Answer 8

I got f'=0 @ x=8 (maximum)
& end point @ x=-1 = > abs min

Answer 9

-1and 8 is wrong

Answer 10

shoot... let me check...
I'll be back in 20 min or no dinner for me

Answer 11

x=-3 as min (for given interval) and
x=8 max looks right - see graph attached

Answer 12

unless I got original equation wrong... did I?
talk to you in 20

Answer 13

u got the equation right

Answer 14

the answer should be right too - see the graph...
max @ x=8 and min @x=-3 (on given interval)