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
Calculate f'(x) to find local extrema:
e^(-x) can never be zero, therefore
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
where u=(3x-1), v=e^(-x)
f'(x)=u'v + v'u
=-e^(-x)[-3 + 3x -1]
Are you OK with the rest?
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