(A) f (x) is differentiable on [−3, −1] so MVT guarantees that there exists
at least one x value such that solving this
equation on your graphing calculator yields x = −2.800, −1.772 on [−3,
−1].(C) See figure below.(D) At t = 0, R(0) = 3,3 and v(0) = 0,0. Given a(t) = 2, e−t, using theFTC