Question

For the function $ f(x) = x^2 e^{-x} $, the maximum occurs when $ x $ is equal to

• A. 2
• B. 1
• C. -1
• D. 0

Hint:

To find the maximum or minimum of a function, take its derivative, set it to zero, and solve for \( x \). Check the second derivative to confirm if it's a maximum or minimum.

Answer 1

The Correct Option is A

To find the value of \( x \) where the function \( f(x) = x^2 e^{-x} \) attains its maximum, we first need to take the derivative of the function with respect to \( x \), set it equal to zero, and solve for \( x \). First, compute the derivative of \( f(x) \): \[ f'(x) = \frac{d}{dx} \left( x^2 e^{-x} \right) \] Using the product rule: \[ f'(x) = 2x e^{-x} - x^2 e^{-x} \] Simplifying: \[ f'(x) = e^{-x} (2x - x^2) \] To find the critical points, set \( f'(x) = 0 \): \[ e^{-x} (2x - x^2) = 0 \] Since \( e^{-x} \neq 0 \) for any real value of \( x \), we solve: \[ 2x - x^2 = 0 \] \[ x(2 - x) = 0 \] This gives \( x = 0 \) or \( x = 2 \). By applying the second derivative test or analyzing the first derivative, it can be shown that the function reaches its maximum at \( x = 2 \).
Therefore, the correct answer is 1. 2.