Question

Let the absolute maxima/minima value of $f\left(x\right)=3x^4-8x^3+12x^2-48x+25;x\in [0,3]$ be max, min. Find $2\left(max\right)+min$?

Answer 1

Given, $f\left(x\right)=3x^4-8x^3+12x^2-48x+25;x\in [0,3]$

$f^{\prime }\left(x\right)=12x^3-24x^2+24x-48=12(x^3-2x^2+2x-4)=12(x-2)(x^2+2)$

For max. or min. of $f\left(x\right)$

$f^{\prime }\left(x\right)=0\Rightarrow x=2$

Now $f^{\prime \prime }\left(x\right)=12(3x^2-4x+2)>\forall x\in R$

Thus $x=2$ is minima of $f\left(x\right)$

$\Rightarrow y_{min}=-39$

$y\left(0\right)=25,y\left(3\right)=16$

Hence in the given interval $y_{max}=25$

$2\left(max\right)+min=2\left(25\right)-39=11$