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.
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
4 answers
How did you get (3x-4) from (3x-1) when you took the first derivative?
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?
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?
Hi MathMate, I have a question, where did the -e^(-x) come from in 3e^(-x) -e^(-x)(3x-1), whereas there is no '-' sign in f'(x)=u'v + v'u?
Thanks.
Thanks.
2 answers