Find both the maximum and the minimum value of `3x^4-8x^3+` `12 x^2-48 x+25` on the interval `[0,3]` .

Explanation

Let `f(x)`
`3x^4-8x^3+`
`12 x^2-48 x+1`

`f^prime (x)=(d(3x^4-8x^3+12x^2-48x+25))/(dx)`

`f^prime (x)=3**4x^3-8**3x^2+12**2x-48+0`

`f^prime (x)=12x^3-24x^2+24x-48
`

`f^prime (x)=12(x^3-2x^2+2x-4)`

...