Consider the function f(x)=33x2\displaystyle {f{{\left({x}\right)}}}={3}-{3}{x}^{{2}} on the interval [3,7]\displaystyle {\left[-{3},{7}\right]}. Find the average or mean slope of the function on this interval, i.e.
f(7)f(3)7(3)=\displaystyle {\frac{{{f{{\left({7}\right)}}}-{f{{\left(-{3}\right)}}}}}{{{7}-{\left(-{3}\right)}}}}=  

By the Mean Value Theorem, we know there exists a c\displaystyle {c} in the open interval (3,7)\displaystyle {\left(-{3},{7}\right)} such that f(c)\displaystyle {f}'{\left({c}\right)} is equal to this mean slope. For this problem, there is only one c\displaystyle {c} that works. Find it.