The absolute minimum value of f(x)=(x^{3}/3)-(5x^{2}/2)-6x on [-1,3] is?

Given,
${\displaystyle f\left(x\right)=\frac{x^3}{3}-\frac{5x^2}{2}-6x,[-1,3]}$
Differentiating with respect to x, we get
${\displaystyle f^{\prime }\left(x\right)=\frac{1}{3}\left(3x^2\right)-\frac{5}{2}\left(2x\right)-6\left(1\right)}$
${\displaystyle =x^2-5x-6}$
Now to find critical values, solve the first derivative by equating it to zero. That is,
f'(x)=0
${\displaystyle \Rightarrow x^2-5x-6=0}$
${\displaystyle \Rightarrow x^2-6x+x-6=0}$
${\displaystyle \Rightarrow x(x-6)+(x-6)=0}$
${\displaystyle \Rightarrow (x-6)(x=1)=0}$
${\displaystyle \Rightarrow x=6\text{or}x=-1}$
But ${\displaystyle 6not\in }$ [-1,3], therefore critical value is x=-1.
Step 2
Now we find the value of the function at the critical value and at the end points of the given interval.
When x=-1,
${\displaystyle f(-1)=\frac{{(-1)}^3}{3}-\frac{5{(-1)}^2}{2}-6(-1)}$
${\displaystyle =-\frac{1}{3}-\frac{5}{2}+6}$
${\displaystyle =\frac{-2-15+36}{6}}$
${\displaystyle =\frac{19}{6}}$
When x=3,
${\displaystyle f\left(3\right)=\frac{{\left(3\right)}^3}{3}-\frac{5{\left(3\right)}^2}{2}-6\left(3\right)}$
${\displaystyle =9-\frac{45}{2}-18}$
${\displaystyle =\frac{18-45-36}{2}}$
${\displaystyle =-\frac{63}{2}}$
Hence absolute minimum value is ${\displaystyle -\frac{63}{2}}$ which occurs at x=3.