Post a New Question

Calculus

posted by .

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 -

    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 -

    How did you get (3x-4) from (3x-1) when you took the first derivative?

  • Calculus -

    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?

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question