Find maximum and minimum value of `y=x^3-9x^2+24x+2`
Solution:
Here, `y=x^3-9x^2+24x+2`
`:. (dy)/(dx)=``=x^(3)-9x^(2)+24x+2`
`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)-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+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=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 maximum
Now, `y=x^3-9x^2+24x+2`
`"putting " x=4`
`y_(min)``=4^(3)-9*4^(2)+24*4+2`
`=64-144+96+2`
`=18`
`"putting " x=2`
`y_(max)``=2^(3)-9*2^(2)+24*2+2`
`=8-36+48+2`
`=22`
This material is intended as a summary. Use your textbook for detail explanation.
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