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
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.
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?
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.
To find the largest and smallest values of the function f(x)=(3x-1)e^-x over the interval 0 < x < 2, we can use calculus.
Step 1: Find the critical points of the function.
To find the critical points, we need to identify the values of x where the derivative of the function is equal to zero or undefined. So, let's differentiate the function with respect to x.
f'(x) = (3 - 3x)e^-x - (3x - 1)e^-x = (3 - 3x - 3x + 1)e^-x = (4 - 6x)e^-x
Setting the derivative equal to zero, we have:
(4 - 6x)e^-x = 0
Since e^-x is always positive, we can ignore it for finding the critical points. Therefore,
4 - 6x = 0
6x = 4
x = 4/6
x = 2/3
Step 2: Determine the endpoints of the interval.
The given interval is 0 < x < 2, so the endpoints are x = 0 and x = 2.
Step 3: Evaluate the function at the critical points and endpoints.
To find the largest and smallest values, we need to compare the values of the function at the critical points and endpoints. So, calculate f(x) for each of these values.
f(0) = (3(0) - 1)e^0 = -1
f(2/3) = (3(2/3) - 1)e^(-2/3) = (2 - 1)e^(-2/3) = e^(-2/3)
f(2) = (3(2) - 1)e^-2 = (6 - 1)e^-2 = 5e^-2
Step 4: Determine the largest and smallest values.
Compare the values obtained in step 3 to find the largest and smallest values of the function.
The smallest value is f(2) = 5e^-2, and the largest value is f(0) = -1.