Question
Find the maximum value of the function g(x) =
4x3 − 24x2 + 48x – 10 on the set T = {x ∈ R ∣ x2 − 14x + 45 ≤ 0} .Solution
Given, S = {x ∈ ℝ /x2 + 45 ≤ 14x} ∴ x2 + 45 ≤ 14x ⇒ x2 - 14x + 45 ≤ 0 ⇒ (x - 5) (x - 9) ≤ 0 ⇒ x ∈ [5, 9] Now, f(x) = 4x3 − 24x2 + 48x – 10 ⇒ f'(x) = 12x2 - 48x + 48 ⇒ f'(x) = 12(x2 - 4x + 4) = 12 [(x2 - 4x + 4) − 1] = 12(x - 2)2 - 12 ∴ f'(x) > 0 ∀ x ∈ [5, 9] ∴ f(x) is strictly increasing in the interval [5, 9] ∴ Maximum value of f(x) when x ∈ [5, 9] is f(9) = 1394
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