For every real number x neq -1, let f(x) = fr | vedclass

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(A) Given $f(x) = \frac{x}{x+1}$.Calculating the first few terms:$f_1(x) = \frac{x}{x+1}$$f_2(x) = f(f(x)) = \frac{\frac{x}{x+1}}{\frac{x}{x+1} + 1} =

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$\frac{2^n}{1 \cdot 3 \cdot 5 \cdot \ldots \cdot (2n-1)}$

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(A) Given $f(x) = \frac{x}{x+1}$.Calculating the first few terms:$f_1(x) = \frac{x}{x+1}$$f_2(x) = f(f(x)) = \frac{\frac{x}{x+1}}{\frac{x}{x+1} + 1} = \frac{x}{x + x + 1} = \frac{x}{2x+1}$$f_3(x) = f(f_2(x)) = \frac{\frac{x}{2x+1}}{\frac{x}{2x+1} + 1} = \frac{x}{x + 2x + 1} = \frac{x}{3x+1}$By induction, $f_n(x) = \frac{x}{nx+1}$.Evaluating at $x = -2$:$f_n(-2) = \frac{-2}{n(-2)+1} = \frac{-2}{-2n+1} = \frac{2}{2n-1}$.The product is $P = f_1(-2) \cdot f_2(-2) \cdot \ldots \cdot f_n(-2) = \prod_{k=1}^{n} \frac{2}{2k-1} = \frac{2^n}{1 \cdot 3 \cdot 5 \cdot \ldots \cdot (2n-1)}$.

For every real number $x \neq -1$, let $f(x) = \frac{x}{x+1}$. Define $f_1(x) = f(x)$ and for $n \geq 2$, $f_n(x) = f(f_{n-1}(x))$. Then the product $f_1(-2) \cdot f_2(-2) \cdot \ldots \cdot f_n(-2)$ is equal to:

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