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The function f(x)=2x3−15x2+36x+4f(x) = 2x^3-15x^2+36x+4 has local maxima at x=x=

Solution

✅ Correct Option: 4

f′(x)=6x2−30x+36=6(x−2)(x−3)f'(x) = 6x^2-30x+36 = 6(x-2)(x-3), so critical points are x=2x=2 and x=3x=3.

f′f' changes from positive to negative at x=2x=2, giving a local maximum there (and a local minimum at x=3x=3).

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