Question

The greatest value of the function \(f(x) = xe^{-x}\) in \([0, \infty)\), is

• A. 0
• B. \(\frac{1}{e}\)
• C. \(-e\)
• D. \(e\)

Hint:

The function \(xe^{-x}\) increases from 0 to \(1/e\) and then decreases to 0.

Answer 1

The Correct Option is B

To find the greatest value of the function \(f(x) = xe^{-x}\) on the interval \([0, \infty)\), we need to perform the following steps:

1.  Calculate the derivative of the function to find critical points:
   • The function is \(f(x) = xe^{-x}\).
   • Using the product rule for differentiation, where \(u = x\) and \(v = e^{-x}\), we have:
   • \(\frac{d}{dx}[uv] = u'v + uv'\).
   • Calculate \(u' = \frac{d}{dx}[x] = 1\).
   • Calculate \(v' = \frac{d}{dx}[e^{-x}] = -e^{-x}\).
   • The derivative is then: \(f'(x) = e^{-x} + x(-e^{-x}) = e^{-x}(1 - x)\).
2. Set the derivative equal to zero to find critical points:
   • Solve \(e^{-x}(1 - x) = 0\).
   • Since \(e^{-x} \neq 0\) for any real \(x\), we have \(1 - x = 0\), which gives \(x = 1\).
3. Evaluate the function at the critical point and endpoints:
   • At \(x = 0\): \(f(0) = 0 \cdot e^{0} = 0\).
   • At \(x = 1\): \(f(1) = 1 \cdot e^{-1} = \frac{1}{e}\).
   • As \(x \to \infty\), \(xe^{-x} \to 0\).
4. Conclusion:
   • The maximum value of \(f(x)\) on \([0, \infty)\) is \(\frac{1}{e}\), which occurs at \(x = 1\).
   • Therefore, the correct answer is \(\frac{1}{e}\).