If f(x) = frac{4x + 3}{6x - 4}, x neq frac{2} | vedclass

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(A) Step $1$: Calculate $(f \circ f)(x)$.$(f \circ f)(x) = f(f(x)) = \frac{4(\frac{4x+3}{6x-4}) + 3}{6(\frac{4x+3}{6x-4}) - 4}$.Step $2$: Simplify the

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(A) Step $1$: Calculate $(f \circ f)(x)$.$(f \circ f)(x) = f(f(x)) = \frac{4(\frac{4x+3}{6x-4}) + 3}{6(\frac{4x+3}{6x-4}) - 4}$.Step $2$: Simplify the expression.$= \frac{16x + 12 + 18x - 12}{24x + 18 - 24x + 16} = \frac{34x}{34} = x$.Step $3$: Since $(f \circ f)(x) = x$, then $g(x) = x$.Step $4$: The composition $(g \circ g \circ g \circ g \circ g)(x)$ is equivalent to $g(g(g(g(g(x)))))$.Since $g(x) = x$, then $g(g(g(g(g(x))))) = x$.Step $5$: Evaluate at $x = 3$.$(g \circ g \circ g \circ g \circ g)(3) = 3$.

If $f(x) = \frac{4x + 3}{6x - 4}$, $x \neq \frac{2}{3}$ and $(f \circ f)(x) = g(x)$ where $g : R - \{\frac{2}{3}\} \to R - \{\frac{2}{3}\}$, then $(g \circ g \circ g \circ g \circ g)(3) = $

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