Answer :

It is given that f (x) = (x − 1)2 + 3, x [3,1]

f’(x) = 2(x – 1)


Now, f’(x) = 0


2(x-1)


x = 1


Now, we evaluate the value of f at critical point x = 1 and at end points of the interval [-3, 1].


f(1) = (1 - 1)2 + 3 = 0 + 3 = 3


f(-3) = (-3 - 1)2 + 3 = 16 + 3 = 19


Therefore, we have the absolute maximum value of f on [-3, 1] is 19 occurring at x =-3.


And, the absolute minimum value of f on [-3,1] is 3 occurring at x = 1.


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