Let f(x) = (x² − 3)eª. (x² − 3)e. The function has two critical points at x = -3, and x = 1. (a) Use the second derivative test to classify each critical point as local max or local min. (b) Find the global maximum value M and the global minimum value m of the function f(x) on the interval ï € [−4, −2]. (c) Consider the definite integral -2 -2 (x² — 3)e* dx. Use your result from part (b) f(x) dx = fe -4 -4 to find an underestimate and overestimate for the integral. (d) Find the right sum using n=2, i.e., RIGHT(2), for the integral -2 -2 [₁² f(x) dx = [ -4 -4 below. Round your answer to two decimal places. 2 (x² − 3)e dx. Enter this value in the blank