If f(x)=(4x+3)/(6x-4),x (2)/(3) and (f f)(x)=g(x), where g:R-(2)/(3) R-(2)/(3), then (gogo…

Question

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

Accepted Answer

To find $(g\circ g\circ g)(4),$ we first need to understand the composition of $f$ with itself, i.e., $(f\circ f)(x)=f(f(x))=g(x).$ We can then repeatedly apply $g$ to get the given expression.

First, let's calculate $(f\circ f)(x)=g(x):$

$g(x)=(f\circ f)(x)=f(f(x))$

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

To evaluate this expression, we substitute $\frac{4x+3}{6x-4}$ for $x$ in the function $f(x):$

$g(x)=f(\frac{4x+3}{6x-4})=\frac{4(\frac{4x+3}{6x-4})+3}{6(\frac{4x+3}{6x-4})-4}$

Now, we simplify the expression:

$g(x)=\frac{4(4x+3)+3(6x-4)}{6(4x+3)-4(6x-4)}$

$=\frac{16x+12+18x-12}{24x+18-24x+16}$

$=\frac{34x}{34}$

$=x$

So, $g(x)=x$ for all $x$ in the domain of $g$, which is $R-{\frac{2}{3}}$. It's important to note that the domain restriction is preserved through the composition because $f(x)$ has a vertical asymptote at $x=\frac{2}{3}$ which doesn't intersect the graph.

So, $g(x)$ is the identity function on its domain, which means that applying $g$ any number of times will result in the same input for $x$ in the given domain. Hence, we have:

$(g\circ g\circ g\circ g)(4)=g(g(g(g(4))))=g(g(g(4)))=g(g(4))=g(4)=4$

This corresponds to option D, which is $4$.

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