Given function is f(x) = (x – 1)2 + 3
On differentiation we get
⇒ f'(x) = 2(x – 1)
Now, for local minima and local maxima we have f'(x) = 0
2(x – 1) = 0
x = 1
Then, we evaluate of f at critical point x = 1 and at the interval [– 3, 1]
f (1) = (1 – 1)2 + 3 = 3
f (– 3) = (– 3 – 1)2 + 3 = 19
Hence, we can conclude that the absolute maximum value of f on [– 3, 1] is 19 occurring at x = – 3 and the minimum value of f on [– 3, 1] is 3 occurring at x = 1Answered by Sakshi | 1 year ago
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