Animated Solution for Mathematics - Functions: Let $f(x)=\frac{x-1}{x+1},x\in \mathbb{R}-\{0,-1,1\}$. If $f^{n+1}(x)=f(f^n(x))$ for all $n\in \mathbb{N}$, then $f^6(6)+f^7(7)$ is equal to:

We are given the function $f(x)=\frac{x-1}{x+1}$. To avoid tedious algebraic repetition, we must determine the behavior of the function under repeated composition.

Let us calculate the second iteration, $f^2(x)$:

By simplifying the expression using the common denominator $(x+1)$, we obtain:

Now, we calculate the third iteration, $f^3(x)$, by substituting $f^2(x)$ back into $f(x)$:

Multiplying the numerator and denominator by $x$ yields:

Finally, we find the fourth iteration, $f^4(x)$:

Since $f^4(x)=x$, the function is periodic with a period of 4.

Given the period of $4$, we can simplify the required terms using remainders:

For $f^6(6)$, we note that $6\equiv 2(\mathrm{mod}4)$, so $f^6(6)=f^2(6)$:

For $f^7(7)$, we note that $7\equiv 3(\mathrm{mod}4)$, so $f^7(7)=f^3(7)$:

Summing these results gives the final answer:

The final result is $-\frac{3}{2}$.