Question

Determine the coordinates of the absolute maximum and minimum values of the function below for the given interval.

f(x) = 3xe^(-x)

For interval -1 less than or equal to x less than or equal to 3

Answer 1

\[[-1,3]\]

Answer 2

\[f'(x)=?\]

Answer 3

3e^(-x)+3x*-e^(-x)

Answer 4

yes, or
\[3e^{-x}-3xe^{-x}=3e^{-x}~(1-x)\]

Answer 5

Thank you I think this is where I was messing up!

Answer 6

now, critical numbers are those that:

1. where f'(x) is undefined, iFF part of f(x) domain.
2. x-values when f'(x)=0
3. closed interval boundaries.

Answer 7

There are no restrictions for x in f(x) or f'(x).

set f'(x)=0

Answer 8

and solve for x.

Answer 9

\[0=3e^{-x}(1-x)\]

Answer 10

x=?

Answer 11

1

Answer 12

yes

Answer 13

I was messing up getting the critical point out of the derivative thanks a lot :-)

Answer 14

so, your interval boundaries (closed - i.e included in your part of the function) are -1 and 3

so critical numbers are {-1, 1, 3}

Answer 15

now, find f(-1) f(1) and f(3) and tell me which one is the minimum and the maximum

Answer 16

min at f(-1) and max at f(1)

Answer 17

check, are you sure ?

Answer 18

Yeah!

Answer 19

f(x) = 3xe^(-x)
f(3)=3(3)e^(-3)=9/e^3=0.44
f(1)=3(1)e^(-1)=3/e=1.1
f(-1)=3(-1)e^(--1)=-3e^2


yes, I was looking at the derivative... you are right. sorry

Answer 20

so,
min is (-1,f(-1))
max is (1,f(1))

Answer 21

bye!

Answer 22

Thank you a ton!