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

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(D) Given $f(x) = \frac{4x+3}{6x-4}$.First,we find $g(x) = (f \circ f)(x) = f(f(x))$.$g(x) = f\left(\frac{4x+3}{6x-4}\right) = \frac{4\left(\frac{4x+3

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$4$

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(D) Given $f(x) = \frac{4x+3}{6x-4}$.First,we find $g(x) = (f \circ f)(x) = f(f(x))$.$g(x) = f\left(\frac{4x+3}{6x-4}\right) = \frac{4\left(\frac{4x+3}{6x-4}\right) + 3}{6\left(\frac{4x+3}{6x-4}\right) - 4}$.Multiplying the numerator and denominator by $(6x-4)$:$g(x) = \frac{4(4x+3) + 3(6x-4)}{6(4x+3) - 4(6x-4)} = \frac{16x + 12 + 18x - 12}{24x + 18 - 24x + 16} = \frac{34x}{34} = x$.Since $g(x) = x$ for all $x$ in the domain,$g$ is the identity function.Therefore,$(g \circ g \circ g)(4) = g(g(g(4))) = g(g(4)) = g(4) = 4$.

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

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