Consider the function f(:c) = 2:03 9:32 24.1: + 9 on the interval [ 5,9]. The average rate of change of x) on this interval is: f(9)f(5) 9(5) =[62 ]d' The Mean Value Theorem says that there is at least one number c in the open interval ( 5, 9) so that f'(c) equals this average rate of change. Find the value(s) of c that work for f(:z:) on ( 5, 9): %] Let f be continuous on [2, 6] and differentiable on (2, 6). If f(2) = 8 and f'(a:) 2 12 for all 3, what is the smallest possible value for f(6)? 40 X