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The absolute maximum value of the function f(x)=2x3−15x2+36x+7f(x) = 2x^3 - 15x^2 + 36x + 7 on the interval [1,5] is

Solution

✅ Correct Option: 3

f′(x)=6(x−2)(x−3)f'(x) = 6(x-2)(x-3), so critical points are x=2,3x = 2, 3, both inside [1,5][1,5]. Values: f(1)=30f(1) = 30, f(2)=35f(2) = 35, f(3)=34f(3) = 34, f(5)=62f(5) = 62. The absolute maximum is 62.

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