The function given by $f(x)=x \exp \left(-x^{2} / 4\right)$ has exactly one point of inflection with a positive $\mathrm{x}$-coordinate. What is the $\mathrm{x}$-coordinate of the point of inflection?

Note: We are using exp to denote the exponential function with base $e$, so that we have $\exp (u)=e^{u}$ for any $u$.

Enter your answer as a decimal. Round to three decimal places (as needed).

Step 1 ：To find the point of inflection, we need to find the second derivative of the function \( f(x) = x e^{-\frac{x^2}{4}} \) and set it equal to zero.

Step 2 ：The second derivative of the function is \( f''(x) = \frac{x(x^2 - 6)}{4} e^{-\frac{x^2}{4}} \).

Step 3 ：We set the second derivative equal to zero and solve for the x-coordinate: \( \frac{x(x^2 - 6)}{4} e^{-\frac{x^2}{4}} = 0 \).

Step 4 ：The solutions to this equation are \( x = 0 \), \( x = -\sqrt{6} \), and \( x = \sqrt{6} \).

Step 5 ：Since we are looking for the point of inflection with a positive x-coordinate, we take \( x = \sqrt{6} \).

Step 6 ：The x-coordinate of the point of inflection is approximately \( 2.449 \) when rounded to three decimal places.

Step 7 ：The final answer is \( \boxed{2.449} \)