Consider the function $\displaystyle {f{{\left({x}\right)}}}={6}-{3}{x}^{{2}}$ on the interval $\displaystyle {\left[-{2},{3}\right]}$. Find the average or mean slope of the function on this interval, i.e.
$\displaystyle {\frac{{{f{{\left({3}\right)}}}-{f{{\left(-{2}\right)}}}}}{{{3}-{\left(-{2}\right)}}}}=$ Preview Question 6 Part 1 of 2  

By the Mean Value Theorem, we know there exists a $\displaystyle {c}$ in the open interval $\displaystyle {\left(-{2},{3}\right)}$ such that $\displaystyle {f}'{\left({c}\right)}$ is equal to this mean slope. For this problem, there is only one $\displaystyle {c}$ that works. Find it.
Preview Question 6 Part 2 of 2