If $f(x) = \dfrac{x - 2011}{11}$, then $(f \circ f \circ f \circ f \circ f)(x)$ is $\cdots$
A. $\frac{x+2011}{x-1}$ B. $\frac{x+2011}{x+1}$ C. $\frac{x-2011}{x+1}$ D. $\frac{x-2011}{x-1}$ E. $\frac{-x+2011}{x-1}$
I find no leading to solve this one. Please help me. I thought there is a sort of pattern so that we can find the faster way for it, but it looks like the first composition already makes the thing too complicated. Can you give me some tips?
Hint: The given answers are incorrect. As for solving the problem, why not try a general case and see if any patterns emerge?
$$f(x) = \frac{x\color{blue}{+b}}{\color{purple}{c}}$$
$$(f\circ f)(x) = \frac{\frac{x+b}{c}+b}{c} = \frac{\frac{x+b+bc}{c}}{c} = \frac{x\color{blue}{+b+bc}}{\color{purple}{c^2}}$$
$$(f\circ f\circ f)(x) = \frac{\frac{x+b+bc}{c^2}+b}{c} = \frac{\frac{x+b+bc+bc^2}{c^2}}{c} = \frac{x\color{blue}{+b+bc+bc^2}}{\color{purple}{c^3}}$$
Notice the geometric progression in the numerators.