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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+24For stationary values, (dy)/(dx)=0=>3x^(2)-18x+24=0=>3(x-2)(x-4)=0=>x-2=0" or "x-4=0=>x=2" or "x=4:.At x=2 and x=4 we get stationary values.Now, (d^2y)/(dx^2)==3x^(2)-18x+24d/(dx)(3x^(2)-18x+24)=d/(dx)(3x^(2))-d/(dx)(18x)+d/(dx)(24)=6x-18+0=6x-18((d^2y)/(dx^2))_(x=2)=6*2-18=12-18=-6 (negative):. At x=2 the function is maximum((d^2y)/(dx^2))_(x=4)=6*4-18=24-18=6 (positive):. At x=4 the function is minimumNow, y=x^3-9x^2+24x+2"putting " x=2y_(max)=2^(3)-9*2^(2)+24*2+2=8-36+48+2=22"putting " x=4y_(min)=4^(3)-9*4^(2)+24*4+2=64-144+96+2=18 Solution provided by AtoZmath.com Any wrong solution, solution improvement, feedback then Submit Here Want to know about AtoZmath.com and me