Consider the function $f(x)=1-8x^2$ on the interval $[-6,7]$. Find the average or mean slope of the function on this interval, i.e.
$$\frac{f(7)-f(-6)}{7-(-6)}=$$
By the Mean Value Theorem, we know there exists a $c$ in the open interval $(-6,7)$ 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 ：Define the function $f(x)=1-8x^2$ on the interval $[-6,7]$.

Step 2 ：Calculate the average slope of the function on this interval using the formula $\frac{f(7)-f(-6)}{7-(-6)}$.

Step 3 ：Substitute $x=7$ and $x=-6$ into the function $f(x)$ to get $f(7)=-391$ and $f(-6)=-287$.

Step 4 ：Substitute these values into the formula for the average slope to get $\frac{-391-(-287)}{7-(-6)}=-8$. So, the average slope of the function on the interval $[-6,7]$ is $-8$.

Step 5 ：Find the derivative of the function $f(x)$, which is $f^{\prime }(x)=-16x$, by applying the power rule for differentiation.

Step 6 ：Set $f^{\prime }(c)$ equal to the average slope and solve for $c$ to find the value of $c$ that makes $f^{\prime }(c)$ equal to the average slope. This gives $c=\frac{1}{2}$.

Step 7 ：Final Answer: The average slope of the function on the interval $[-6,7]$ is $\boxed{{\displaystyle -8}}$. The value of $c$ that makes $f^{\prime }(c)$ equal to the average slope is $\boxed{{\displaystyle \frac{1}{2}}}$.