Consider the function $f(x)=1-8 x^{2}$ 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 \(f'(x) = -16x\), by applying the power rule for differentiation.

Step 6 ：Set \(f'(c)\) equal to the average slope and solve for \(c\) to find the value of \(c\) that makes \(f'(c)\) 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 \(\boxed{-8}\). The value of \(c\) that makes \(f'(c)\) equal to the average slope is \(\boxed{\frac{1}{2}}\).