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Question 72

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

$$f(f(x)) = f\left(\frac{4x+3}{6x-4}\right) = \frac{4 \cdot \frac{4x+3}{6x-4} + 3}{6 \cdot \frac{4x+3}{6x-4} - 4}$$

Simplify the numerator: $$\frac{4(4x+3) + 3(6x-4)}{6x-4} = \frac{16x+12+18x-12}{6x-4} = \frac{34x}{6x-4}$$
Simplify the denominator: $$\frac{6(4x+3) - 4(6x-4)}{6x-4} = \frac{24x+18-24x+16}{6x-4} = \frac{34}{6x-4}$$

Therefore, $$f(f(x)) = \frac{34x/(6x-4)}{34/(6x-4)} = \frac{34x}{34} = x$$ which means that $$g(x) = f(f(x))$$ is the identity function.

Since $$g$$ is the identity function, $$(g \circ g \circ g)(4) = g(g(g(4))) = g(g(4)) = g(4) = 4.$$

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