Find the absolute maximum and absolute minimum values of `f(x)=3x^4-8x^3+12 x^2-48 x+25` in `[0,\ 3]`

Explanation

`f(x)=3x^(4)-8x^(3)+12x^(2)-48x+25`
`f(x)=12x^(3)-24x^(2)+24x-48`
`=12(x^(3)-2x^(2)+2x-4)`
`=12[x^(2)(x-2)+2(x-2)]`
`=12(x-2)(x^(2)+2)`
Now f(x) =0 gives f(x)=2 which is point of minima
Now f(2)=3(16)-8(8)+12(4)-48(2)+25=-38
Also `underset(xrarrPM00)f(x)=00`
Therefore range of the function is [-39,00]