How do you find the maximum or minimum values for $f(x) = x - \left( {\dfrac{{64x}}{{x + 4}}} \right)$ on the interval $[0,13]$ ?

Hint:
For solving this particular problem we need to differentiate the given function with respect to the independent variable that is $x$, then equate the required result to zero. By Solving the equation, we get the values for $x$. Substitute the values in the second derivative of the given function. If the result after substitution is less than zero . Then you have to consider it as the maximum point, and if you get the result greater than zero, you have to consider it as minimum.

Complete step by step solution:
We have
$f(x) = x - \left( {\dfrac{{64x}}{{x + 4}}} \right)$ ,
We have to find the maximum and minimum values on the interval $[0,13]$ .
Therefore, we have to differentiate the given function first ,
We will get ,
$f'(x) = 1 - \left( {\dfrac{{64(x + 4) - 64x}}{{{{(x + 4)}^2}}}} \right)$
           $
   = \dfrac{{{{(x + 4)}^2} - [64x + 256 - 64x]}}{{{{(x + 4)}^2}}} \\
   = \dfrac{{{x^2} + 8x + 16 - 64x - 256 + 64x}}{{{{(x + 4)}^2}}} \\
   = \dfrac{{{x^2} + 8x - 240}}{{{{(x + 4)}^2}}} \\
 $
Now , set $f'(x) = 0$ ,
We will the following result ,
$ \Rightarrow \dfrac{{{x^2} + 8x - 240}}{{{{(x + 4)}^2}}} = 0$
Multiply by ${(x + 4)^2}$ both the side , we will get ,
$ \Rightarrow {x^2} + 8x - 240 = 0$
Solve this by splitting the middle term as ,
$
   \Rightarrow {x^2} + 20x - 12x - 240 = 0 \\
   \Rightarrow x(x + 20) - 12(x + 20) = 0 \\
   \Rightarrow (x - 12)(x + 20) = 0 \\
 $
Therefore, we will get $x = 12$ and $x = - 20$ . so, there are two critical points, $12$ and $ - 20$. Since $ - 20$ is not in the interval we care about, we drop it from our list. Adding the endpoints to the list, we have
        $0 < 12 < 13$
as our ordered list of special points.
Now double differentiate the given function as,
$f''(x) = 2x + 8$
$f''(12) = 2(12) + 8$
           $ = 22$
Here, $f''(12) > 0$ , this will give minimum value ,
$f(12) = 12 - \left( {\dfrac{{64 \times 12}}{{12 + 4}}} \right) = - 36$
Therefore, it is the minimum value.

Note:
A function $f(x)$ encompasses a local maximum or relative maximum at $x$ equals to ${x_0}$ if the graph of $f(x)$near ${x_0}$ features a peak at ${x_0}$. A function $f(x)$ features a local minimum or relative minimum at $x$ equals to ${x_0}$ if the graph of $f(x)$ near ${x_0}$ encompasses a trough at ${x_0}$. (To make the excellence clear, sometimes the ‘plain’ maximum and minimum are called absolute maximum and minimum.)