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Find maximum and minimum value of y=x^3-9x^2+24x+2

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Your problem `->` maximum and minimum value of y=x^3-9x^2+24x+2


Here, `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`

For 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+24`

`d/(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 minimum

Now, `y=x^3-9x^2+24x+2`

`"putting " x=2`

`y_(max)``=2^(3)-9*2^(2)+24*2+2`

`=8-36+48+2`

`=22`

`"putting " x=4`

`y_(min)``=4^(3)-9*4^(2)+24*4+2`

`=64-144+96+2`

`=18`






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