Part 1 of 5 To find the absolute maximum and minimum values for f(t)=16-22 on the interval [-1, 4], we must find the largest and smallest function values at the critical numbers and at the interval endpoints. First, we'll find f'(t). Part 2 of 5 f'(t)=1 (16-2)-(1)(~ (-21)) + (16-2) (3) (1) ( (16-17) - (-21)) + + (16-2)* (1) After simplifying, we have f'(t) = 16 - 2t 16-12 The critical points occur where this is 0 or undefined. The fraction will be O when the numerator is 0. This occurs at Part 3 of 5 t= 22 22 16 - 2t2 The function f'(t) = 16-12 will be undefined when the denominator is 0. This happens when t = 4 4 Part 4 of 5 Note that t = -8 and t = -4 are not in the interval. Now find the following function values. f(-1): = f(8) = f(4) =