Consider the function $f(x)=4 x^{3}-3 x$ on the interval $[-3,3]$. Find the average or mean slope of the function on this interval.

By the Mean Value Theorem, we know there exists at least one $c$ in the open interval $(-3,3)$ such that $f^{\prime}(c)$ is equal to this mean slope.
For this problem, there are two values of $c$ that work.
The smaller one is
and the larger one is

Step 1 ：Consider the function \(f(x)=4 x^{3}-3 x\) on the interval \([-3,3]\). 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 at least one \(c\) in the open interval \((-3,3)\) such that \(f^{\prime}(c)\) is equal to this mean slope.

Step 3 ：To find the mean slope, we first need to calculate the function values at \(x=-3\) and \(x=3\). Then we subtract the function value at \(x=-3\) from the function value at \(x=3\), and divide by the length of the interval, which is \(3 - (-3) = 6\).

Step 4 ：The function values at \(x=-3\) and \(x=3\) are \(-99\) and \(99\) respectively. So, the mean slope is \((99 - (-99))/6 = 33\).

Step 5 ：After finding the mean slope, we can set the derivative of the function equal to this mean slope and solve for \(c\) to find the values of \(c\) that satisfy the Mean Value Theorem.

Step 6 ：The derivative of the function is \(f^{\prime}(x) = 12x^{2} - 3\). Setting this equal to the mean slope, we get \(12x^{2} - 3 = 33\). Solving this equation gives us the values of \(c\) as \(-\sqrt{3}\) and \(\sqrt{3}\).

Step 7 ：Final Answer: The mean slope of the function on the interval \([-3,3]\) is \(\boxed{33}\). The values of \(c\) that satisfy the Mean Value Theorem are \(\boxed{-\sqrt{3}}\) and \(\boxed{\sqrt{3}}\).