The function f(x) = 2 x^3 - 15 x^2 + 36x + 4 is maximum at x = ________________ .

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2 	ext { Given }: f\left( x ight) = 2 x^3 - 15 x^2 + 36x + 4 \Rightarrow f'\left( x ight) = 6 x^2 - 30x + 36 	ext { For a local maxima or a local minima, we must have } f'\left( x ight) = 0 \Rightarrow 6 x^2 - 30x + 36 = 0 \Rightarrow x^2 - 5x + 6 = 0 \Rightarrow \left( x - 2 ight)\left( x - 3 ight) = 0 \Rightarrow x = 2, 3 	ext { Now }, f''\left( x ight) = 12x - 30 \Rightarrow f''\left( 2 ight) = 24 - 30 = - 6 < 0 	ext { So, x = 1 is a local maxima } . 	ext { Also }, f''\left( 3 ight) = 36 - 30 = 6 > 0 	ext { So,x = 2 is a local maxima }.

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