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 Solution Find maximum and minimum value of y=x^3-9x^2+24x+2Solution:Your problem -> maximum and minimum value of y=x^3-9x^2+24x+2Here, y=x^3-9x^2+24x+2:. (dy)/(dx)=d/(dx)(x^(3)-9x^(2)+24x+2)=d/(dx)(x^(3))-d/(dx)(9x^(2))+d/(dx)(24x)+d/(dx)(2)=3x^(2)-18x+24+0=3x^(2)-18x+24=3(x^(2)-6x+8)=3(x-2)(x-4)For stationary values, (dy)/(dx)=0=>3(x-2)(x-4)=0=>3(x-2)(x-4)=0=3(x^(2)-6x+8)=>x=0:.At x=0 we get stationary values.Now, (d^2y)/(dx^2)=d/(dx)(3(x-2)(x-4))=3(d/(dx)(x-2))(x-4)+3(x-2)(d/(dx)(x-4))d/(dx)(x-2)=1d/(dx)(x-2)=d/(dx)(x)-d/(dx)(2)=1-0=1d/(dx)(x-4)=1d/(dx)(x-4)=d/(dx)(x)-d/(dx)(4)=1-0=1=3(1)(x-4)+3(x-2)(1)=3(x-4)+3(x-2)=3x-12+3x-6=6x-18=6(x-3)((d^2y)/(dx^2))_(x=0)=6(0-3)=6(-3)=-18 (negative):. At x=0 the function is maximumNow, y=x^3-9x^2+24x+2"putting " x=0y_(max)=(0)^(3)-9(0)^(2)+24(0)+2=0-0+0+2=2 Solution provided by AtoZmath.com Any wrong solution, solution improvement, feedback then Submit Here Want to know about AtoZmath.com and me