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, \(f'(x)\), equal to the mean slope and solve for \(x\). The derivative of \(f(x) = \frac{1}{x}\) is \(f'(x) = -\frac{1}{x^2}\).

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

Step 6 ：Final Answer: \(c = \boxed{\sqrt{7}}\)