Question 987872
<font face="Times New Roman" size="+2">


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  f(x)\ =\ x^3\ +\ 11]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \frac{f(x)\ -\ f(a)}{x\ -\ a}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \frac{(x^3\ +\ 11)\ -\ (a^3\ +\ 11)}{x\ -\ a]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \frac{x^3\ -\ a^3}{x\ -\ a}]


Recall the factorization of the difference of two cubes:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \frac{(x\ -\ a)(x^2\ +\ ax\ +\ a^2)}{x\ -\ a}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  x^2\ +\ ax\ +\ a^2]


Now you can take the limit as x approaches a:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \lim_{x\right{a}}\,x^2\ +\ ax\ +\ a^2\ =\ a^2\ +\ a^2\ +\ a^2\ =\ 3a^2]


Therefore


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ M_a\ =\ 3a^2]


Hence, at *[tex \Large a\ =\ 3]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ M_{a=3}\ =\ 3(3)^2\ =\ 27]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \