To find the greatest value of the function \( f(x) = xe^{-x} \) in the interval \([0, \infty)\), we will apply calculus techniques such as differentiation.
- \(f(x) = xe^{-x}\) is the function under consideration.
- First, we find the critical points by differentiating \( f(x) \) and setting the derivative equal to zero. The derivative of \( f(x) \) is found using the product rule:
\[ f'(x) = \frac{d}{dx} (xe^{-x}) = e^{-x} + x(-e^{-x}) \] \(f'(x) = e^{-x} - xe^{-x} = e^{-x}(1-x)\). - Set \( f'(x) \) to zero to find the critical points: \[ e^{-x}(1-x) = 0 \] Since \( e^{-x} \neq 0 \) for all \( x \), we have: \[ 1-x = 0 \quad \Rightarrow \quad x = 1 \]
- Evaluate \( f(x) \) at the critical point and at the endpoint of the interval \([0, \infty)\).
- At \( x = 1 \): \[ f(1) = 1 \times e^{-1} = \frac{1}{e} \]
- At \( x = 0 \): \[ f(0) = 0 \times e^{0} = 0 \]
- For behavior as \( x \to \infty \): \[ \lim_{x \to \infty} f(x) = \lim_{x \to \infty} xe^{-x} = 0 \] because \( e^{-x} \) approaches 0 faster than \( x \) grows.
Thus, the maximum value of \( f(x) \) in the interval \([0, \infty)\) occurs at \( x = 1 \) and is \(\frac{1}{e}\).
Therefore, the correct answer is \(\frac{1}{e}\).