Consider the function $f(x)=4x^3-3x$ 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 $f(x)=4x^3-3x$ 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 $\boxed{{\displaystyle 33}}$. The values of $c$ that satisfy the Mean Value Theorem are $\boxed{{\displaystyle -\sqrt{3}}}$ and $\boxed{{\displaystyle \sqrt{3}}}$.