Let [f(x) = x - 1 x + 1,,x R - 0, - 1,1 ].

Question

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

Accepted Answer

Given,

$$f(x) = {{x - 1} \over {x + 1}}$$

Also given,

${f^{n + 1}}(x) = f({f^n}(x))$ ..... (1)

$$\therefore$$ For $n = 1$

${f^{1 + 1}}(x) = f({f^1}(x))$

$$ \Rightarrow {f^2}(x) = f(f(x))$$

$$ = f\left( {{{x - 1} \over {x + 1}}} \right)$$

$$ = {{{{x - 1} \over {x + 1}} - 1} \over {{{x - 1} \over {x + 1}} + 1}}$$

$$ = {{{{x - 1 - x - 1} \over {x + 1}}} \over {{{x - 1 + x + 1} \over {x + 1}}}}$$

$$ = {{ - 2} \over {2x}}$$

$$ =  - {1 \over x}$$

From equation (1), when n = 2

${f^{2 + 1}}(x) = f({f^2}(x))$

$$ \Rightarrow {f^3}(x) = f({f^2}(x))$$

$$ = f\left( { - {1 \over x}} \right)$$

$$ = {{ - {1 \over x} - 1} \over { - {1 \over x} + 1}}$$

$$ = {{{{ - 1 - x} \over x}} \over {{{ - 1 + x} \over x}}}$$

$$ = {{ - 1 - x} \over { - 1 + x}} = {{ - (x + 1)} \over {x - 1}}$$

Similarly,

${f^4}(x) = f({f^3}(x))$

$$ = f\left( {{{ - x + 1} \over {x - 1}}} \right)$$

$$ = {{{{ - (x + 1)} \over {x - 1}} - 1} \over {{{ - (x + 1)} \over {x - 1}} + 1}}$$

$$ = {{{{ - x - 1 - x + 1} \over {x - 1}}} \over {{{ - x - 1 + x - 1} \over {x - 1}}}}$$

$$ = {{ - 2x} \over { - 2}} = x$$

$$\therefore$$ ${f^5}(x) = f({f^4}(x))$

$= f(x)$

$$ = {{x - 1} \over {x + 1}}$$

${f^6}(x) = f({f^5}(x))$

$$ = f\left( {{{x - 1} \over {x + 1}}} \right)$$

$$ =  - {1 \over x}$$ (Already calculated earlier)

${f^7}(x) = f({f^6}(x))$

$$ = f\left( { - {1 \over x}} \right)$$

$$ = {{ - {1 \over x} - 1} \over { - {1 \over x} + 1}}$$

$$ = {{ - (x + 1)} \over {x - 1}}$$

$$\therefore$$ $${f^6}(6) =  - {1 \over 6}$$

and $${f^7}(7) = {{ - (7 + 1)} \over {7 - 1}} =  - {8 \over 6}$$

So, ${f^6}(6) + {f^7}(7)$

$$ =  - {1 \over 6} - {8 \over 6}$$

$$ =  - {3 \over 2}$$

Solutions · 1