Question

The function \( f(x) = x e^{-x^2} \) has a minimum at

• A. \( x = \sqrt{2} \)
• B. \( x = -\sqrt{2} \)
• C. \( x = \frac{1}{\sqrt{2}} \)
• D. \( x = -\frac{1}{\sqrt{2}} \)

Hint:

To find the minimum of a function, differentiate, set the derivative equal to zero, and use the second derivative test to confirm the nature of the critical point.

Answer 1

The Correct Option is D

Step 1: Differentiate the function.
The function is \( f(x) = x e^{-x^2} \). To find the minimum, we need to take the derivative of the function and set it equal to zero. Using the product rule for differentiation:
\[ f'(x) = \frac{d}{dx}(x) \cdot e^{-x^2} + x \cdot \frac{d}{dx}(e^{-x^2}) = e^{-x^2} - 2x^2 e^{-x^2} \] Step 2: Set the derivative equal to zero.
To find critical points, we set \( f'(x) = 0 \):
\[ e^{-x^2} (1 - 2x^2) = 0 \] Since \( e^{-x^2} \) is never zero, we have:
\[ 1 - 2x^2 = 0 \implies x^2 = \frac{1}{2} \implies x = \pm \frac{1}{\sqrt{2}} \] Step 3: Second derivative test.
To confirm that this is a minimum, we check the second derivative:
\[ f''(x) = -2x e^{-x^2} + 4x^3 e^{-x^2} - 2x e^{-x^2} = e^{-x^2}(4x^3 - 4x) \] Substituting \( x = \frac{1}{\sqrt{2}} \) into the second derivative gives a positive value, confirming it is a minimum.
Step 4: Conclusion.
Thus, the function has a minimum at \( x = \frac{1}{\sqrt{2}} \), which corresponds to option (C).