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

Explanation

Let `f(x) = 3x^(4) - 8x^(3)+12x^(2)-48x+25`
` rArr 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)`
For maxima/minima, f'(x) = 0,
` rArr 12(x-2)(x^(2)+2)=0`
` rArr x-2=0 rArr x = 2 in [0, 3]`
and ` x^(2)+2 ne 0`
Now, we find the values of f(x) at x = 2 and at the end points of the interval [0, 3].
at ` x = 2, f(2) = 3 xx 2^(4) - 8 xx 2^(3) + 12 xx 2^(2) - 48 xx 2 + 25`
`= 48 - 64 + 48 - 96 + 25 =- 39`
at `x = 0, f(0)=0-0+0-+25=25`
at ` x = 3 , f(3) = 3 xx 3^(4) - 8 xx 3^(3) + 12 xx 3^(2) - 48 xx 3 + 25`
` = 243 - 216 + 108 - 144 + 25 = 16`
`:. f(x) ` has absolute maximum value 25 at x = 0 and absolute minimum value- 39 at x = 2.