Question

Let (f) be the function defined by (f(x)=xe^(1-x)) for all real numbers (x). Find the range of (f).

Answer 1

The range of the function f(x) = xe^(1-x) for the given interval (0 ≤ x ≤ 20) is [0, 1.10364].

Step-by-step explanation:

The range of a function is the set of all possible output values. To find the range of the function f(x) = xe^(1-x), we need to determine the set of possible values that f(x) can take on.

1. To start, we can analyze the behavior of the function as x approaches positive infinity and negative infinity. As x approaches positive infinity, both x and e^(1-x) become very large, resulting in f(x) also becoming very large. Similarly, as x approaches negative infinity, both x and e^(1-x) become very small, resulting in f(x) approaching zero.
2. Next, we can consider the behavior of the function within the given interval (0 ≤ x ≤ 20). By evaluating the function at the endpoints of the interval, we can determine the minimum and maximum values of f(x) within the given interval. Plugging in x = 0, we get f(0) = 0. Multiplying any value of x in the interval by e^(1-x) will also be positive, so the minimum value of f(x) is 0.
3. To find the maximum value, we need to consider the derivative of f(x). Taking the derivative of f(x) with respect to x, we get f'(x) = e^(1-x) - xe^(1-x). Setting f'(x) equal to zero and solving for x, we find that the maximum occurs at x = 2. Plugging in x = 2, we get f(2) = 2e^(-1) ≈ 1.10364.

Therefore, the range of the function f(x) = xe^(1-x) for the given interval (0 ≤ x ≤ 20) is [0, 1.10364].