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 Problem: maximum and minimum value of y=x^3-9x^2+24x+2 [ Calculator, Method and examples ]Solution: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^2-6x+8)=0=>3(x^2-2x-4x+8)=0=>3(x(x-2)-4(x-2))=0=>3(x-4)(x-2)=0=>x-4=0" or "x-2=0=>x=4" or "x=2:.At x=4 and x=2 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=4)=6*(4)-18=24-18=6 (positive):. At x=4 the function is minimum((d^2y)/(dx^2))_(x=2)=6*(2)-18=12-18=-6 (negative):. At x=2 the function is maximumNow, y=x^3-9x^2+24x+2"putting " x=4y_(min)=(4)^3-9*(4)^2+24*(4)+2=64-144+96+2=18"putting " x=2y_(max)=(2)^3-9*(2)^2+24*(2)+2=8-36+48+2=22

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