Question: f(x) = xe^{-x^2/162}, [-7, 18] absolute minimum value absolute maximum value

f(x) = xe^{-x^2/162}, [-7, 18] absolute minimum value absolute maximum value

Solution

To find the absolute minimum and maximum values of the function \( f(x) = xe^{-x^2/162} \) on the interval \([-7, 18]\), we first need to find the critical points and evaluate the function at these points as well as at the endpoints of the interval. Find the derivative of \( f(x) \). Using the product rule for differentiation, where \( u = x \) and \( v = e^{-x^2/162} \), we have: \[ f’(x) = u’v + uv’ \] Compute \( u’ \) and \( v’ \): \( u’ = 1 \) \( v’ = e^{-x^2/162} \cdot \left(-\frac{2x}{162}\right) = -\frac{2x}{162} e^{-x^2/162} \) Now substitute into the product rule formula: \[ f’(x) = 1 \cdot e^{-x^2/162} + x \left(-\frac{2x}{162} e^{-x^2/162}\right) \] Simplify the expression: \[ f’(x) = e^{-x^2/162} - \frac{2x^2}{162} e^{-x^2/162} \] \[ f’(x) = e^{-x^2/162} \left(1 - \frac{2x^2}{162}\right) \] Set \( f’(x) = 0 \): \[ e^{-x^2/162} \left(1 - \frac{2x^2}{162}\right) = 0 \] Since \( e^{-x^2/162} \neq 0 \), simplify: \[ 1 - \frac{2x^2}{162} = 0 \] \[ 1 = \frac{2x^2}{162} \] \[ 2x^2 = 162 \] \[ x^2 = 81 \] \[ x = \pm 9 \] Evaluate \( f(x) \) at critical points and endpoints: 1. \( f(-7) = -7e^{-(-7)^2/162} \) 2. \( f(9) = 9e^{-9^2/162} = 9e^{-81/162} = 9e^{-1/2} \) 3. \( f(-9) = -9e^{-(-9)^2/162} = -9e^{-1/2} \) 4. \( f(18) = 18e^{-(18)^2/162} = 18e^{-324/162} = 18e^{-2} \) Compare values: - \( f(-7) = -7e^{-49/162} \) - \( f(9) = 9e^{-1/2} \) - \( f(-9) = -9e^{-1/2} \) - \( f(18) = 18e^{-2} \) Thus, the absolute minimum value is \( f(-9) = -9e^{-1/2} \). The absolute maximum value is \( f(9) = 9e^{-1/2} \).