Step-by-step Solution:

The function is given by \( f(x) = |x+1|e^{-x^2} \). To find critical points, we analyze the domains \( x > -1 \) and \( x < -1 \). Case 1: \( x > -1 \) Here, \( |x+1| = x+1 \), so \( f(x) = (x+1)e^{-x^2} \). Differentiate \( f(x) \) with respect to \( x \): \[ f'(x) = 1 \cdot e^{-x^2} + (x+1)e^{-x^2}(-2x) \] \[ f'(x) = e^{-x^2} [1 - 2x(x+1)] = e^{-x^2} (-2x^2 - 2x + 1) \] Set \( f'(x) = 0 \) to find critical points: \[ -2x^2 - 2x + 1 = 0 \implies 2x^2 + 2x - 1 = 0 \] Using the quadratic formula: \[ x = \frac{-2 \pm \sqrt{4 - 4(2)(-1)}}{4} = \frac{-2 \pm \sqrt{12}}{4} = \frac{-1 \pm \sqrt{3}}{2} \] This gives roots \( x_1 \approx 0.366 \) and \( x_2 \approx -1.366 \). Since we are in the region \( x > -1 \), the valid critical point is \( x = \frac{-1 + \sqrt{3}}{2} \approx 0.366 \). Case 2: \( x < -1 \) Here, \( |x+1| = -(x+1) \), so \( f(x) = -(x+1)e^{-x^2} \). Differentiate \( f(x) \) with respect to \( x \): \[ f'(x) = e^{-x^2} (2x^2 + 2x - 1) \] Set \( f'(x) = 0 \) to find critical points. The roots are the same: \( x = \frac{-1 \pm \sqrt{3}}{2} \). Since we are in the region \( x < -1 \), the valid critical point is \( x = \frac{-1 - \sqrt{3}}{2} \approx -1.366 \). Now, analyze the nature of the critical point at \( x = \frac{-1 - \sqrt{3}}{2} \): For \( x < -1.366 \), \( 2x^2 + 2x - 1 > 0 \), so \( f'(x) > 0 \) (function is increasing). For \( -1.366 < x < -1 \), \( 2x^2 + 2x - 1 < 0 \), so \( f'(x) < 0 \) (function is decreasing). Since \( f'(x) \) changes from positive to negative, \( x = \frac{-1 - \sqrt{3}}{2} \) is a point of local maximum. Notice that \( -2 < \frac{-1 - \sqrt{3}}{2} < -1 \). Therefore, \( f \) has a point of maxima in the interval \( (-2, -1) \).