To solve this problem, let's apply the Mean Value Theorem (MVT). The MVT states that if a function $f(x)$ is continuous on the closed interval $[a, b]$ and differentiable on the open interval $(a, b)$, then there exists at least one $c$ in $(a, b)$ such that:

$f'(c) = \frac{f(b) - f(a)}{b - a}$

Step 1: Identify the Function and Interval

The given function is: $f(x) = 9x + \frac{9}{x}$ The interval is $\left[\frac{1}{19}, 19\right]$, where $a = \frac{1}{19}$ and $b = 19$.

Step 2: Compute $f(a)$ and $f(b)$

1. Calculate $f\left(\frac{1}{19}\right)$: $f\left(\frac{1}{19}\right) = 9 \cdot \frac{1}{19} + \frac{9}{\frac{1}{19}} = \frac{9}{19} + 9 \cdot 19 = \frac{9}{19} + 171$

2. Calculate $f(19)$: $f(19) = 9 \cdot 19 + \frac{9}{19} = 171 + \frac{9}{19}$

Step 3: Compute the Average Rate of Change $\frac{f(b) - f(a)}{b - a}$

The average rate of change of $f(x)$ on $\left[\frac{1}{19}, 19\right]$ is: $\frac{f(b) - f(a)}{b - a} = \frac{\left(171 + \frac{9}{19}\right) - \left(\frac{9}{19} + 171\right)}{19 - \frac{1}{19}} = \frac{0}{19 - \frac{1}{19}}$

So