Find the absolute maximum and minimum values of f(x)=(x^2)(e^(-x/2)) on the interval [-1,6]
http://tutorial.math.lamar.edu/Classes/CalcI/AbsExtrema.aspx
Thanks so much
To find the absolute maximum and minimum values of a function on a closed interval, we need to examine the critical points and the endpoints of the interval.
1. Find the critical points by setting the derivative of the function f(x) equal to zero and solving for x.
First, let's find the derivative of f(x):
f'(x) = 2x(e^(-x/2)) + (x^2)(-e^(-x/2))(1/2)
Setting f'(x) equal to zero:
2x(e^(-x/2)) + (x^2)(-e^(-x/2))(1/2) = 0
Simplifying the equation, we get:
2x(e^(-x/2)) - (x^2)(e^(-x/2))(1/2) = 0
Factoring out e^(-x/2), we have:
(e^(-x/2))(2x - x^2/2) = 0
To find critical points, we solve the equation:
e^(-x/2)(2x - x^2/2) = 0
Either e^(-x/2) = 0 (which has no solution) or 2x - x^2/2 = 0
Solving 2x - x^2/2 = 0, we get:
x(4 - x) = 0
x = 0 or x = 4
So, the critical points are x = 0 and x = 4.
2. Evaluate the function at the critical points and the endpoints of the interval [-1, 6].
f(-1) = (-1)^2 * e^(-(-1)/2) = e^(1/2) = 1.6487 (approximately)
f(0) = (0)^2 * e^(-0/2) = 0
f(4) = (4)^2 * e^(-4/2) = 8e^-2 = 0.1099 (approximately)
f(6) = (6)^2 * e^(-6/2) = 36e^-3 = 0.1341 (approximately)
3. Compare the values of f(x) at the critical points and endpoints to determine the absolute maximum and minimum.
The maximum value is the largest of the values f(-1) = 1.6487, f(0) = 0, f(4) = 0.1099, and f(6) = 0.1341.
The minimum value is the smallest of the values f(-1) = 1.6487, f(0) = 0, f(4) = 0.1099, and f(6) = 0.1341.
Therefore, the absolute maximum value of f(x) on the interval [-1,6] is approximately 1.6487, and the absolute minimum value is approximately 0.