Find the absolute max and min values at the indicated interval f(x)=2x^3-x^2-4x+10 ,[-1,0]

Given,
${\displaystyle f\left(x\right)=2x^3-x^2-4x+10,[-1,0]}$
Absolute maximum and absolute minimum values of a function exist at a point where its first derivative is zero or at the end points of the given interval.
So first we find the derivative of the given function:
${\displaystyle f^{\prime }\left(x\right)=2\cdot 3x^2-2x-4+0}$
${\displaystyle \Rightarrow f^{\prime }\left(x\right)=6x^2-2x-4}$
Now f'(x)=0
${\displaystyle \Rightarrow 6x^2-2x-4=0}$
${\displaystyle \Rightarrow 6x^2-6x+4x-4=0}$
${\displaystyle \Rightarrow 6x(x-1)+4(x-1)=0}$
${\displaystyle \Rightarrow (x-1)(6x+4)=0}$
${\displaystyle \Rightarrow x-1=0\text{or}6x+4=0}$
${\displaystyle \Rightarrow x=1\text{or}x=-\frac{2}{3}}$
Now computing the value of the function at these points and at the end points of the interval:
at ${\displaystyle x=-1,f(-1)=f\left(x\right)=2{(-1)}^3-{(-1)}^2-4(-1)+10=11}$
at ${\displaystyle x=0,f\left(0\right)=f\left(x\right)=2{\left(0\right)}^3-{\left(0\right)}^2-4\left(0\right)+10=10}$
at ${\displaystyle x=-\frac{2}{3},f(-\frac{2}{3})=f\left(x\right)=2{(-\frac{2}{3})}^3-{(-\frac{2}{3})}^2-4(-\frac{2}{3})+10=12.8148}$
We didn't find the value of f(1), because 1 does not belongs to the interval ${\displaystyle [-1,0]}$.
Hence absolute maximum value is 12.8148 and absolute minimum value is 10.