Find the absolute extrema of
f(x)= (2x-1)e^(-x) on [0, 00)
f ' (x) = (2x-1) e^(-x) ( -1) + 2 e^(-x)
= 0 for a max/min
e^(-x) (2 - (2x-1)) = 0
e^(-x) ( 3 - 2x) = 0
e^(-x) = 0 ----> no solution
or
3 - 2x = 0
x = 3/2
then f(3/2) = 2e^(-3/2) = 2/e^(3/2)
To find the absolute extrema of a function on a given interval, we need to find both the maximum and minimum values of the function within that interval. In this case, we are looking for the absolute extrema of the function f(x) = (2x-1)e^(-x) on the interval [0, ∞).
To find the absolute extrema, we need to consider two types of points:
1. Critical points: These are the points where the derivative of the function is either zero or undefined. We need to check the critical points within the interval [0, ∞).
2. Endpoints: These are the points at the boundary of the interval.
Let's start by finding the critical points. To do that, we need to find the derivative of f(x) and solve for x where the derivative is zero or undefined.
f(x) = (2x-1)e^(-x)
Take the derivative of f(x) with respect to x:
f'(x) = (2e^(-x)) + (-2x + 1)e^(-x)
Now, set f'(x) equal to zero and solve for x:
(2e^(-x)) + (-2x + 1)e^(-x) = 0
Factoring out e^(-x):
e^(-x)(2 - 2x + 1) = 0
Simplifying:
2 - 2x + 1 = 0
-2x + 3 = 0
-2x = -3
x = 3/2
So, x = 3/2 is a critical point.
Next, let's consider the endpoints of the interval, which are 0 and ∞.
Now, we can evaluate the function at the critical points and endpoints to find the maximum and minimum values:
Evaluate f(x) at x = 0:
f(0) = (2(0)-1)e^(-0) = -1
Evaluate f(x) at x = 3/2:
f(3/2) = (2(3/2)-1)e^(-3/2) = 1.5e^(-3/2)
Now, we need to analyze f(x) as x approaches ∞. As x gets larger, the value of e^(-x) gets closer and closer to zero. Therefore, f(x) approaches zero as x approaches ∞.
So, the minimum value of f(x) on the interval [0, ∞) is 0, and the maximum value is the larger of the two values we calculated: -1 and 1.5e^(-3/2).
To compare these values, we need to determine if 1.5e^(-3/2) is less than or greater than -1.
To make this comparison, we can approximate e^(-3/2) using a calculator:
e^(-3/2) ≈ 0.22313
So, 1.5e^(-3/2) ≈ 1.5 * 0.22313 ≈ 0.3347
Comparing this value with -1, we see that -1 is less than 0.3347.
Therefore, the maximum value of f(x) on the interval [0, ∞) is 0.3347, and the minimum value is -1.
To summarize:
Absolute maximum value: 0.3347 (approximately)
Absolute minimum value: -1