f(x)=2x3−15x2+36x+10→(i)differentiatew.rtox⇒f′(x)=6x2−30x+36⇒f′(x)=0⇒6x2−30x+36=0⇒x2−5x+6=0⇒x2−3x−2x+6=0⇒x(x−3)−2(x−3)=0⇒x=3,x=2againdifferentiatew.rtoxf′(x)=12x−30→(ii)Putx=2⇒f′(2)=24−30=−60,x=2pointmaximumandminimumvalueisf(2)=2×23−15×22+36×2+10⇒f(2)=16−60+71+10=38andputx=3ineqn(ii)⇒f′(3)=36−30=6Itispointmaximumandminimumvalueisf(3)=2×27−15×9+36×3+10=37