Let f(x) = ₀^ x^2 - 2x (t^2 + t + ) e^ -t dt , where and are real constants. If the function f(x) has local maxima at x=0 and x=2 , and local minima at x=-1 , x=1 , and x=3 , then the ordered pair ( , ) is equal to

Step-by-step solution

Given f(x) = ₀^ x^2 - 2x (t^2 + t + ) e^ -t dt Using the Leibniz rule for differentiation under the integral sign, we get: f'(x) = ( (x^2-2x)^2 + (x^2-2x) + ) e^ -(x^2-2x) d dx (x^2-2x) f'(x) = ( (x^2-2x)^2 + (x^2-2x) + ) e^ -(x^2-2x) 2(x-1) For critical points, f'(x) = 0 . Since e^ -(x^2-2x) > 0 for all real x , the critical points are given by: x - 1 = 0 x = 1 and (x^2-2x)^2 + (x^2-2x) + = 0 We are given that the local extrema occur at x = -1, 0, 1, 2, 3 . The point x=1 is already accounted for by the factor 2(x-