The average cost function associated with producing and marketing x units of a given item is given by $\text{AC}=2x-11+\dfrac{50}{x}$ . Find the range of values of the output x, for which AC is increasing.

Hint: In this question, we are given the average cost of producing and marketing x units of the item as the function of x as $\text{AC}=2x-11+\dfrac{50}{x}$. We know that the value of a function will be increasing if its derivative is positive. Therefore, we should take the derivative of the given function, and find the values of x, for which the derivative is positive. This range will be the answer to this question.

Complete step by step solution:
The average cost function associated with producing and marketing x units of a given item is given as $\text{AC}=2x-11+\dfrac{50}{x}.......................(1.1)$
 We have to find the range of x for which the average cost is increasing. We know a function will be increasing if its derivative is positive. Therefore, we should find the derivative of AC and check the range for which the derivative is positive…………………..(1.1a)
We know the following rules for derivatives
$\begin{align}
  & \dfrac{d\left( f(x)+g(x) \right)}{dx}=\dfrac{df(x)}{dx}+\dfrac{dg(x)}{dx}................(1.2) \\
 & \dfrac{d\left( cf(x) \right)}{dx}=c\dfrac{df(x)}{dx}...........................(1.3) \\
 & \dfrac{d\left( \dfrac{1}{x} \right)}{dx}=\dfrac{-1}{{{x}^{2}}}..............................(1.4) \\
 & \dfrac{dx}{dx}=1.............................(1.5) \\
 & \dfrac{dc}{dx}=0...............................(1.6) \\
\end{align}$
Where $f(x)$ and $g(x)$ are functions of x and c is any constant.
Therefore, using equations (1.2), (1.3) , (1.4), (1.5) and (1.6) in equation (1.1), we obtain
$\begin{align}
  & \dfrac{d\left( \text{AC} \right)}{dx}=\dfrac{d\left( 2x-11+\dfrac{50}{x} \right)}{dx}=2\dfrac{dx}{dx}-\dfrac{d\left( 11 \right)}{dx}+50\dfrac{d\left( \dfrac{1}{x} \right)}{dx} \\
 & =2\times 1-0+50\times \dfrac{-1}{{{x}^{2}}}=2-\dfrac{50}{{{x}^{2}}}.....................(1.7) \\
\end{align}$
Now, from (1.1a), as the derivative should be positive for AC to be increasing, from (1.7), we can write
$\begin{align}
  & \dfrac{d\left( \text{AC} \right)}{dx}>0 \\
 & \Rightarrow 2-\dfrac{50}{{{x}^{2}}}>0\Rightarrow 2>\dfrac{50}{{{x}^{2}}}\Rightarrow {{x}^{2}}>\dfrac{50}{2}=25 \\
 & \Rightarrow {{x}^{2}}>{{5}^{2}}............................(1.8) \\
\end{align}$
Now, we see that the square of a number will be greater than the square of another number if its magnitude is greater than that of the other number. Therefore, from (1.8), we can write
$\begin{align}
  & {{x}^{2}}>{{5}^{2}} \\
 & \Rightarrow \left| x \right|>\left| 5 \right| \\
 & \Rightarrow x\in (-\infty ,-5)\cup (5,\infty )....................(1.9) \\
\end{align}$
which is the required answer to this question.

Note: We should note that in equation (1.9), we should take the negative values whose magnitude is greater than 5 in the answer as there is no restriction on x to be positive or negative. Also, we could have expressed the same answer as $\{x:x<-5\text{ or }x>5\}$ as the answer in the set theoretic way. However, both representations represent the same range of x and hence the same answer.