(1 point) Consider the function $f(x)=2 x^{3}-9 x^{2}-60 x+7$ on the interval $[-6,7]$. Find the average or mean slope of the function on this interval.

Average slope:

By the Mean Value Theorem, we know there exists at least one value $c$ in the open interval $(-6,7)$ such that $f^{\prime}(c)$ is equal to this mean slope. List all values $c$ that work. If there are none, enter none.

Values of $c$ :

Step 1 ：Given the function \(f(x)=2 x^{3}-9 x^{2}-60 x+7\) on the interval \([-6,7]\).

Step 2 ：We need to find the average or mean slope of the function on this interval.

Step 3 ：The average slope of a function on an interval [a, b] is given by the formula \((f(b) - f(a)) / (b - a)\).

Step 4 ：Calculate this for the given function and interval to get the average slope.

Step 5 ：After calculating the average slope, we need to find the values of 'c' for which the derivative of the function equals the average slope.

Step 6 ：This can be done by setting the derivative of the function equal to the average slope and solving for 'c'.

Step 7 ：The derivative of the function \(f(x)\) is \(f'(x) = 6x^{2} - 18x - 60\).

Step 8 ：Setting this equal to the average slope and solving for 'c' gives us the values \(c = \frac{3}{2} - \frac{\sqrt{543}}{6}\) and \(c = \frac{3}{2} + \frac{\sqrt{543}}{6}\).

Step 9 ：Final Answer: The average slope of the function on the interval \([-6,7]\) is \(\boxed{17}\). The values of \(c\) that satisfy the Mean Value Theorem are \(\boxed{\frac{3}{2} - \frac{\sqrt{543}}{6}}\) and \(\boxed{\frac{3}{2} + \frac{\sqrt{543}}{6}}\).