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?
Answer this Question
- Another Calculus - Sorry I have been trying to figure this out all day. Find the...
- Calculus Help!! - Region R is bounded by the functions f(x) = 2(x-4) + pi, g(x...
- Calculus IVT - The function f is continuous on the closed interval [0,6] and ...
- Calculus - 1. Locate the absolute extrema of the function f(x)=cos(pi*x) on the ...
- Calculus - The function f(x)=(x^4)-(10x^3)+(18x^2)-8 is continuous on the closed...
- Calculus HELP!!! - Here is the graph. h t t p : / /goo.gl/PTc2I (spaces added at...
- Calculus - Find the absolute maximum and minimum values of f on the given closed...
- mathematics - find the smallest and largest values of the function...
- calculus - The function f is continuous on the closed interval [0,2] and has ...
- calculus - The function f is continuous on the closed interval [0,6] and has ...