Consider the function $f(x)=\frac{1}{x}$ on the interval $[1,7]$. Find the average or mean slope of the function on this interval.

By the Mean Value Theorem, we know there exists a $c$ in the open interval $(1,7)$ such that $f^{\prime }(c)$ is equal to this mean slope. For this problem, there is only one $c$ that works. Find it.

Step 1 ：Consider the function $f(x)=\frac{1}{x}$ on the interval $[1,7]$. We are asked to find the average or mean slope of the function on this interval.

Step 2 ：By the Mean Value Theorem, we know there exists a $c$ in the open interval $(1,7)$ such that $f^{\prime }(c)$ is equal to this mean slope. For this problem, there is only one $c$ that works. We need to find it.

Step 3 ：The mean slope of a function on an interval $[a,b]$ is given by the difference in the function values at the endpoints of the interval divided by the length of the interval. In this case, the function is $f(x)=\frac{1}{x}$, and the interval is $[1,7]$. So, the mean slope is $\frac{f(7)-f(1)}{7-1}$.

Step 4 ：To find the value of $c$ that satisfies the Mean Value Theorem, we need to set the derivative of the function, $f^{\prime }(x)$, equal to the mean slope and solve for $x$. The derivative of $f(x)=\frac{1}{x}$ is $f^{\prime }(x)=-\frac{1}{x^2}$.

Step 5 ：The solution to the equation $f^{\prime }(x)=\text{mean\_slope}$ is $x=\pm \sqrt{7}$. However, only the positive solution is in the interval $(1,7)$, so $c=\sqrt{7}$.

Step 6 ：Final Answer: $c=\boxed{{\displaystyle \sqrt{7}}}$