The absolute maximum value of the function $f\left(x\right)=2x^3-9x^2+12x+1$ on $\left[0, 2\right]$ is

The absolute maximum value of the function $f\left(x\right)=2x^3-9x^2+12x+1$ on $\left[0, 2\right]$ 

Given, $f\left(x\right)=2x^3-9x^2+12x+1...................\left(i\right)$

Differentiating w. r. t. $x$

$f^{\text{'}}\left(x\right)=6x^2-18x+12$

$f^{\text{'}}\left(x\right)=0$

$6x^2-18x+12=0$

$x^2-3x+2=0$

$x^2-2x-x+2=0$

$x\left(x-2\right)-1\left(x-2\right)=0$

$\left(x-2\right)\left(x-1\right)=0$

$x=2$ and $x=1$

Putting $x=2$ in equation $\left(i\right)$

$f\left(2\right)=2{\left(2\right)}^3-9{\left(2\right)}^2+12\left(2\right)+1$

$f\left(2\right)=5$

Putting $x=1$ in equation $\left(i\right)$

$f\left(1\right)=2{\left(1\right)}^3-9{\left(1\right)}^2+12\left(1\right)+1$

$f\left(1\right)=6$

Now, also we need to check the values of the function at $x=0$

$f\left(0\right)=2{\left(0\right)}^3-9{\left(0\right)}^2+12\left(0\right)+1$

$f\left(0\right)=1$

Hence, The absolute maximum value of function is $6$.