given that y=(3x+1)^2+(2x-1)^3 find the points on the curve for which dy/dx=0

Dear Student,

y=3x+12+2x-13dydx=2(3x+1)×3 +3(2x-1)×2      =6(3x+1)+6(2x-1)      =6(3x+1+2x-1)      =6(5x)      =30xfor dydx=030x=0implies x=0Now,at x=0,the value of y would be:y=3x+12+2x-13y=12+(-1)3=1-1=0Therefore,the only point where dydx=0 is (0,0) 


Regards

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