Posted by **Carly** on Sunday, July 10, 2011 at 2:59pm.

Find the largest and smallest values of the given function over the prescribed closed, bounded interval of:

f(x)=(3x-1)e^-x for 0<x<2

- Calculus -
**MathMate**, Sunday, July 10, 2011 at 5:42pm
f(x)=(3x-1)e^-x

Calculate f'(x) to find local extrema:

f'(x)=-(3x-4)e^(-x)

e^(-x) can never be zero, therefore

(3x-4)=0, or

x=4/3, which is on the interval [0,2].

f"(4/3)=-0.79 (4/3 is a maximum)

Now calculate f(0) and f(2) and compare the three values

f(0),f(2) and f(4/3) to determine which are the maximum and minimum.

- Calculus -
**Carly**, Monday, July 11, 2011 at 2:57pm
How did you get (3x-4) from (3x-1) when you took the first derivative?

- Calculus -
**MathMate**, Monday, July 11, 2011 at 4:09pm
If we take

f(x)=(3x-1)e^(-x)

=u.v

where u=(3x-1), v=e^(-x)

then

f'(x)=u'v + v'u

=3e^(-x) -e^(-x)(3x-1)

=-e^(-x)[-3 + 3x -1]

=-e^(-x)(3x-4)

Are you OK with the rest?

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