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 Solution Find maximum and minimum value of y=4x^3+19x^2-14x+3Solution:Your problem -> maximum and minimum value of y=4x^3+19x^2-14x+3Here, y=4x^3+19x^2-14x+3:. (dy)/(dx)==(4x^(3)+19x^(2)-14x+3)=4x^(3)+19x^(2)-14x+3d/(dx)(4x^(3)+19x^(2)-14x+3)=d/(dx)(4x^(3))+d/(dx)(19x^(2))-d/(dx)(14x)+d/(dx)(3)=12x^(2)+38x-14+0=12x^(2)+38x-14=2(6x^(2)+19x-7)=2((3x-1)(2x+7))=2(3x-1)(2x+7)For stationary values, (dy)/(dx)=0=>2(3x-1)(2x+7)=0=2(6x^(2)+19x-7)=>x=0:.At x=0 we get stationary values.Now, (d^2y)/(dx^2)=d/(dx)(2(3x-1)(2x+7))=2(d/(dx)(3x-1))(2x+7)+2(3x-1)(d/(dx)(2x+7))d/(dx)(3x-1)=3d/(dx)(3x-1)=d/(dx)(3x)-d/(dx)(1)=3-0=3d/(dx)(2x+7)=2d/(dx)(2x+7)=d/(dx)(2x)+d/(dx)(7)=2+0=2=2(3)(2x+7)+2(3x-1)(2)=6(2x+7)+4(3x-1)=12x+42+12x-4=24x+38=2x(12+19/x)=2(12x+19)=24x+38((d^2y)/(dx^2))_(x=0)=24(0)+38=0+38=38 (positive):. At x=0 the function is minimumNow, y=4x^3+19x^2-14x+3"putting " x=0y_(min)=4(0)^(3)+19(0)^(2)-14(0)+3=0+0+0+3=3 Solution provided by AtoZmath.com Any wrong solution, solution improvement, feedback then Submit Here Want to know about AtoZmath.com and me