Question

The function \( f(x) = xe^{-x} \) for all \( x \in \mathbb{R} \) attains a maximum value at \( x = k \), then \( k = \)

• A. \(1 \)
• B. \(2 \)
• C. \(\frac{1}{e} \)
• D. \(3 \)

Hint:

To find the maximum or minimum value of a function \(f(x)\): 1. Find the first derivative \(f'(x)\). 2. Set \(f'(x) = 0\) and solve for \(x\) to find the critical points. 3. Use the second derivative test: If \(f''(x)<0\) at a critical point, it's a local maximum. If \(f''(x)>0\) at a critical point, it's a local minimum. If \(f''(x) = 0\), the test is inconclusive, and further analysis (e.g., first derivative test) is needed.

Answer 1

The Correct Option is A

Step 1: Find the first derivative of the function.
The given function is \(f(x) = xe^{-x}\).
To find the maximum value of the function, we need to find its critical points by setting the first derivative equal to zero. We will use the product rule for differentiation, which states that if \(f(x) = u(x)v(x)\), then \(f'(x) = u'(x)v(x) + u(x)v'(x)\). Let \(u(x) = x\) and \(v(x) = e^{-x}\).
Then, \(u'(x) = \frac{d}{dx}(x) = 1\).
And, \(v'(x) = \frac{d}{dx}(e^{-x}) = e^{-x} \cdot \frac{d}{dx}(-x) = -e^{-x}\).
Now, apply the product rule to find \(f'(x)\): \[ f'(x) = (1)(e^{-x}) + (x)(-e^{-x}) \] \[ f'(x) = e^{-x} - xe^{-x} \] Factor out \(e^{-x}\): \[ f'(x) = e^{-x}(1 - x) \] Step 2: Find the critical points by setting the first derivative to zero.
To find the critical points, set \(f'(x) = 0\): \[ e^{-x}(1 - x) = 0 \] Since \(e^{-x}\) is always positive for any real value of \(x\) (i.e., \(e^{-x} \neq 0\)), we must have: \[ 1 - x = 0 \] \[ x = 1 \] So, the critical point is \(x = 1\). This is the value of \(k\). 
Step 3: Use the second derivative test to confirm that it is a maximum.
To confirm that \(x = 1\) corresponds to a maximum value, we can use the second derivative test. Find the second derivative \(f''(x)\).
From \(f'(x) = e^{-x}(1 - x)\), again use the product rule.
Let \(u(x) = e^{-x}\) and \(v(x) = 1 - x\).
Then, \(u'(x) = -e^{-x}\).
And, \(v'(x) = -1\).
Now, apply the product rule to find \(f''(x)\): \[ f''(x) = (-e^{-x})(1 - x) + (e^{-x})(-1) \] \[ f''(x) = -e^{-x} + xe^{-x} - e^{-x} \] \[ f''(x) = xe^{-x} - 2e^{-x} \] Factor out \(e^{-x}\): \[ f''(x) = e^{-x}(x - 2) \] Now, evaluate \(f''(x)\) at the critical point \(x = 1\): \[ f''(1) = e^{-1}(1 - 2) \] \[ f''(1) = e^{-1}(-1) \] \[ f''(1) = -\frac{1}{e} \] Since \(f''(1) = -\frac{1}{e}<0\), the function \(f(x)\) attains a local maximum value at \(x = 1\). 
Therefore, the value of \(k\) is 1. The final answer is $\boxed{1}$.