Question

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

Answer 1

Lst $f\left(x\right)=3x^4-8x^3+12x^2-48x+25$
$\Rightarrow f^{\prime }\left(x\right)=12x^3-24x^2+24x-48=12(x^3-2x^2-4)=12\left(x^2\right(x-2)+2(x-2\left)\right)=12(x-2)(x^2+2)$
For maxima or minima put f'(x)=0
$\Rightarrow 12(x-2)(x^2+2)=0\Rightarrow ifx-2=0\Rightarrow x=2ϵ[0,3]$
and if $x^2+2=0\to x^2=-2\Rightarrow x=\sqrt{-2}$
Hence, the only real root is x=2 which is considered as critical point.
Now, we evaluate the value of f at critical point x=2 and at the end points of the interval [0,3].
$\begin{array}{cc}Atx=2& f\left(2\right)=3\times 2^4-8\times 2^3+12\times 2^2-48\times 2+25\\ =-39Atx=0& f\left(0\right)=0-0+0-0+25=25\\ Atx=3,& f\left(3\right)=3\times 3^4-8\times 3^3+12\times 3^2-48\times 3+25\\ & =243-216+108-144+25=16\end{array}$
Hence, we can conclude that the absolute maximum value of f is 25 at x=0 and the absolute minimum value of f is -39 at x=2