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

Step 1
Given:
Given that function ${\displaystyle f\left(x\right)=5+54x-2x^3}$
The given interval is ${\displaystyle [0,4]}$
To find:
we have to find the absolute maximum and absolute minimum
Step 2
Here ${\displaystyle f\left(x\right)=5+54x-2x^3}$ ...(1) and the interval is ${\displaystyle [0,4]}$
Differentiate (1) w.r.t.x, we get
${\displaystyle \frac{d}{dx}f\left(x\right)=\frac{d}{dx}(5+54x-2x^3)}$
${\displaystyle f^{\prime }\left(x\right)=54-6x^2}$
set $f^{\prime }(x)=0$
${\displaystyle \Rightarrow 54-6x^2=0}$
${\displaystyle \Rightarrow 6x^2=54}$
${\displaystyle \Rightarrow x^2=9}$
${\displaystyle \Rightarrow x=3}$ and $x=-3$
but ${\displaystyle x=-3\notin [0,4]}$
and it also contain the critical point $x=0$
Step 3
Hence, the critical points is $x=3$ and $x=0$
Now, ${\displaystyle f\left(x\right)=5+54x-2x^3}$
For $x=3$
${\displaystyle f\left(3\right)=5+54\left(3\right)-2{\left(3\right)}^3=5+162-54}$
$=113$
Thus, the absolute value is maximum at $x=3$
Now, ${\displaystyle f\left(x\right)=5+54x-2x^3}$
For $x=0$
$f(0)=5$
Thus, the absolute value is minimum at $x=0$