Find maximum and minimum value of `y=x^3+6x^2-15x+7`

Solution:
Here, `y=x^3+6x^2-15x+7`

`:. (dy)/(dx)``d/(dx)(x^(3)+6x^(2)-15x+7)`

`=d/(dx)(x^(3))+d/(dx)(6x^(2))-d/(dx)(15x)+d/(dx)(7)`

`=3x^(2)+12x-15+0`

`=3x^(2)+12x-15`

For stationary values, `(dy)/(dx)=0`

`=>3x^(2)+12x-15=0`

`=>3(x^(2)+4x-5)=0`

`=>3(x^(2)-x+5x-5)=0`

`=>3(x(x-1)+5(x-1))=0`

`=>3(x+5)(x-1)=0`

`=>x+5=0" or "x-1=0`

`=>x=-5" or "x=1`

`:.`At `x=-5` and `x=1` we get stationary values.

Now, `(d^2y)/(dx^2)``d/(dx)(3x^(2)+12x-15)`

`=d/(dx)(3x^(2))+d/(dx)(12x)-d/(dx)(15)`

`=6x+12-0`

`=6x+12+0`

`=6x+12`

`((d^2y)/(dx^2))_(x=-5)``=6(-5)+12`

`=-30+12`

`=-18` (negative)

`:.` At `x=-5` the function is maximum

`((d^2y)/(dx^2))_(x=1)``=6*1+12`

`=6+12`

`=18` (positive)

`:.` At `x=1` the function is minimum

Now, `y=x^3+6x^2-15x+7`

`"putting " x=-5`

`y_(max)``=(-5)^(3)+6(-5)^(2)-15(-5)+7`

`=-125+150+75+7`

`=107`

`"putting " x=1`

`y_(min)``=1^(3)+6*1^(2)-15*1+7`

`=1+6-15+7`

`=-1`