Let f(x) = frac{x+1}{x-1} for all x neq 1. Le | vedclass

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(B) Given $f(x) = \frac{x+1}{x-1}$.First,calculate $f^2(x) = f(f(x)) = \frac{\frac{x+1}{x-1} + 1}{\frac{x+1}{x-1} - 1} = \frac{x+1+x-1}{x+1-(x-1)} = \

Vedclass · Accepted Answer

$375$

Vedclass · Answer

(B) Given $f(x) = \frac{x+1}{x-1}$.First,calculate $f^2(x) = f(f(x)) = \frac{\frac{x+1}{x-1} + 1}{\frac{x+1}{x-1} - 1} = \frac{x+1+x-1}{x+1-(x-1)} = \frac{2x}{2} = x$.Since $f^2(x) = x$,we have $f^3(x) = f(f^2(x)) = f(x)$ and $f^4(x) = f^2(f^2(x)) = x$.Now,calculate the terms:$f^1(2) = \frac{2+1}{2-1} = 3$.$f^2(3) = 3$ (since $f^2(x) = x$).$f^3(4) = f(4) = \frac{4+1}{4-1} = \frac{5}{3}$.$f^4(5) = 5$ (since $f^4(x) = x$).Thus,$P = 3 \cdot 3 \cdot \frac{5}{3} \cdot 5 = 75$.The multiples of $75$ are $75, 150, 225, 300, 375, \dots$.Comparing with the given options,$375$ is a multiple of $75$.

Let $f(x) = \frac{x+1}{x-1}$ for all $x \neq 1$. Let $f^1(x) = f(x)$,$f^2(x) = f(f(x))$ and generally $f^n(x) = f(f^{n-1}(x))$ for $n > 1$. Let $P = f^1(2) \cdot f^2(3) \cdot f^3(4) \cdot f^4(5)$. Which of the following is a multiple of $P$?

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