Find the absolute maximum and absolute minimum values of f

Find the absolute maximum and absolute minimum values of f on the given interval.
$f\left(x\right)=2{x}^{3}-3{x}^{2}-12x+1,\left[-2,3\right]$
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Step 1
The function is given by
$f\left(x\right)=2{x}^{3}-3{x}^{2}-12x+1,\left[-2,3\right]$
To evaluate : The absolute maximum and minimum of the function on the given interval.
Step 2
Let us evaluate the critical numbers,
$\frac{df}{dx}=0$
$⇒\frac{d}{dx}\left(2{x}^{3}-3{x}^{2}-12x+1\right)=0$
$⇒6{x}^{2}-6x-12=0$
$⇒{x}^{2}-x-2=0$
$⇒{x}^{2}-2x+x-2=0$
$⇒x\left(x-2\right)+1\left(x-2\right)=0$
$⇒\left(x+1\right)\left(x-2\right)=0$
$⇒x=-1,2$ These are critical numbers
Step 3
Now, let us evaluate the values of the function at the critical numbers and at the endpoints of the closed interval [-2,3]
At x=-1,
$f\left(-1\right)=2×{\left(-1\right)}^{3}-3×{\left(-1\right)}^{2}-12×\left(-1\right)+1=8$
Absolute maximum
At x=2,
$f\left(2\right)=2×{\left(2\right)}^{3}-3×{\left(2\right)}^{2}-12×\left(2\right)+1=-19$
Absolute minimum
At x=-2,
$f\left(-2\right)=2×{\left(-2\right)}^{3}-3×{\left(-2\right)}^{2}-12×\left(-2\right)+1=-3$
At x=3,
$f\left(3\right)=2×{\left(3\right)}^{3}-3×{\left(3\right)}^{2}-12×\left(3\right)+1=-8$
Step 4
Hence, the absolute maximum and absolute minimum values of the function are :
Absolute maximum value of the function = 8
Absolute minimum value of the function =-19