Find the absolute maximum and absolute minimum values of f on the given interval. f(x)=5+54x-2x^3, x in[0,4]

Given,
${\displaystyle f\left(x\right)=5+54x-2x^3,x\in [0,4]}$
Absolute maximum or absolute minimum values of a function on the given interval exists at a point at which its first derivative is zero or at the end points of the given interval.
So differentiating given function with respect to x, we get
${\displaystyle f^{\prime }\left(x\right)=0+54\left(1\right)-2\left(3x^2\right)}$
${\displaystyle \Rightarrow f^{\prime }\left(x\right)=54-6x^2}$
Now f'(x)=0 gives
${\displaystyle 54-6x^2=0}$
${\displaystyle \Rightarrow 6x^2=54}$
${\displaystyle \Rightarrow x^2=9}$
${\displaystyle \Rightarrow x=\pm 3}$
But ${\displaystyle -3\notin [0,4]}$, therefore
x=3
Step 2
Now,
${\displaystyle f\left(0\right)=5+54\left(0\right)-2{\left(0\right)}^3=5}$
${\displaystyle f\left(3\right)=5+54\left(3\right)-2{\left(3\right)}^3=113}$
${\displaystyle f\left(4\right)=5+54\left(4\right)-2{\left(4\right)}^3=93}$
Therefore absolute minimum value is 5 occurs at the point x = 0 and absolute maximum value is 113 occur at the point
x =3.