Find the absolute maximum and absolute minimum values of \( f \) on the given interval. \[ f(x)=x e^{-x^{2} / 50}, \quad[-4,10] \] absolute minimum value \( \square x \)

To find the absolute maximum and minimum values of the function \( f(x)=x e^{-x^{2} / 50} \) on the interval \([-4,10]\), we start by calculating the derivative \( f'(x) \) and setting it to 0 to find critical points. 1. **Finding the derivative**: Using the product rule: \[ f'(x) = e^{-x^{2}/50} + x \left(-\frac{2x}{50} e^{-x^{2}/50}\right) = e^{-x^{2}/50} \left(1 - \frac{2x^2}{50}\right) \] Setting \( f'(x) = 0 \) gives: \[ 1 - \frac{2x^2}{50} = 0 \implies \frac{2x^2}{50} = 1 \implies x^2 = 25 \implies x = 5 \text{ or } x = -5 \] 2. **Evaluating at critical points and endpoints**: Now we evaluate \( f(x) \) at the critical points \( x = -5, 5 \) and at endpoints \( x = -4, 10 \): - \( f(-4) = -4 e^{-4^2/50} = -4 e^{-16/50} \) - \( f(-5) = -5 e^{-5^2/50} = -5 e^{-25/50} \) - \( f(5) = 5 e^{-5^2/50} = 5 e^{-25/50} \) - \( f(10) = 10 e^{-10^2/50} = 10 e^{-100/50} = 10 e^{-2} \) 3. **Finding values**: We can compute these numerically: - \( f(-4) \approx -4 e^{-0.32} \approx -3.57 \) - \( f(-5) \approx -5 e^{-0.5} \approx -3.06 \) - \( f(5) \approx 5 e^{-0.5} \approx 3.06 \) - \( f(10) \approx 10 e^{-2} \approx 1.35 \) 4. **Comparing values**: Comparing all function values: - \( f(-4) \approx -3.57 \) - \( f(-5) \approx -3.06 \) - \( f(5) \approx 3.06 \) - \( f(10) \approx 1.35 \) The absolute minimum value is approximately \( -3.57 \) at \( x = -4 \), and the absolute maximum value is approximately \( 3.06 \) at \( x = 5 \). Thus, the absolute minimum value is \(-3.57\) at \( x = -4 \).