Calculus
posted by Carly .
Find the largest and smallest values of the given function over the prescribed closed, bounded interval of:
f(x)=(3x1)e^x for 0<x<2

f(x)=(3x1)e^x
Calculate f'(x) to find local extrema:
f'(x)=(3x4)e^(x)
e^(x) can never be zero, therefore
(3x4)=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. 
How did you get (3x4) from (3x1) when you took the first derivative?

If we take
f(x)=(3x1)e^(x)
=u.v
where u=(3x1), v=e^(x)
then
f'(x)=u'v + v'u
=3e^(x) e^(x)(3x1)
=e^(x)[3 + 3x 1]
=e^(x)(3x4)
Are you OK with the rest?
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