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 of a function, set its first derivative equal to zero to find critical points, then use the second derivative test to confirm whether it is a maximum.

Answer 1

The Correct Option is B

To find the maximum of the function \( f(x) = x^2 e^{-x} \), we first need to find its derivative with respect to \( x \). This is done by applying the product rule, as the function is a product of two functions: \( x^2 \) and \( e^{-x} \). The derivative of \( f(x) \) is given by: \[ f'(x) = \frac{d}{dx}\left(x^2 e^{-x}\right) \] Using the product rule, we get: \[ f'(x) = \frac{d}{dx}(x^2) \cdot e^{-x} + x^2 \cdot \frac{d}{dx}(e^{-x}) \] \[ f'(x) = 2x \cdot e^{-x} + x^2 \cdot (-e^{-x}) \] \[ f'(x) = e^{-x} (2x - x^2) \] Now, to find the critical points, we set the derivative equal to zero: \[ e^{-x} (2x - x^2) = 0 \] Since \( e^{-x} \) is never zero, we can solve: \[ 2x - x^2 = 0 \] Factor the quadratic equation: \[ x(2 - x) = 0 \] So, \( x = 0 \) or \( x = 2 \). Next, we check the second derivative to determine if \( x = 2 \) is a maximum. We take the second derivative of \( f(x) \): \[ f''(x) = \frac{d}{dx}\left( e^{-x} (2x - x^2) \right) \] Using the product rule again: \[ f''(x) = e^{-x} \left( 2 - 2x \right) - e^{-x} (2x - x^2) \] Simplify: \[ f''(x) = e^{-x} \left( 2 - 4x + x^2 \right) \] Substituting \( x = 2 \) into this second derivative: \[ f''(2) = e^{-2} \left( 2 - 4(2) + 2^2 \right) = e^{-2} \left( 2 - 8 + 4 \right) = e^{-2}(-2) \] Since \( f''(2)<0 \), we confirm that \( x = 2 \) is a maximum point. Thus, the maximum of the function occurs at \( x = 1 \).