Find the ab

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.