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Say $f(x)=dfrac{3x+1}{2}$

$f(f(x))=f^2(x)= dfrac{3(frac{3x+1}{2})+1}{2} =
dfrac{9x+5}{4}$

$f(f(f(x)))=f^3(x)=
dfrac{27x+19}{8}$

I want to predict $f^7(x)$ without manually calculating $f^4(x)$,$f^5(x)$,$f^6(x)$

I clearly can see that
$f^n(x)= dfrac{3^n+something}{2^n}$

How do I find that $something $?

Hint: $$f^n(x) = frac{3^nx+something}{2^n}$$

Do you notice any pattern for $2^n+something$? Notice that you get

$$f(x) = frac{3x+1}{2} implies 1+2 = ?$$

$$f^2(x) = frac{9x+5}{4} implies 5+4 = ?$$

Hint.

$$
f(x) = frac{a x+1}{b}\
f^2(x) = frac{a^2x + frac{a^2-b^2}{a-b}}{b^2}\
vdots\
f^n(x) = frac{a^nx + frac{a^n-b^n}{a-b}}{b^n}
$$

We guess that $$f_n(x)={3^ncdot x+3^n-2^nover 2^n}$$to prove that using induction we have $$f_{n+1}(x)={f(f_n(x))}={3f_n(x)+1over 2}={3^{n+1}cdot x+3^{n+1}-3cdot 2^n+2^nover 2^{n+1} }={3^{n+1}x+3^{n+1}-2^{n+1}over 2^{n+1}}$$since $f_1(x)={3x+1over 2}$ the proof is complete.

Let's solve $f(x)=x$:
$$frac{3x+1}{2}=x$$
$$3x+1=2x$$
$$x=-1$$
So
$$f(-1)=-1$$
This means that
$$f^2(-1)=f(f(-1))=f(-1)=-1$$
and so on.
So based on your conjecture:
$$f^n(x)=frac{3^nx+c(n)}{2^n}$$
Implies that
$$frac{-3^n+c(n)}{2^n}=-1$$
$$-3^n+c(n)=-2^n$$
$$c(n)=3^n-2^n$$
So
$$f^n(x)=frac{3^nx+(3^n-2^n)}{2^n}$$
And why is your conjecture true?
Because
$$f(x)=frac{3}{2}x+c_1$$
$$f^2(x)=frac{3}{2}left(frac{3}{2}x+c_1right)+c_1=frac{3^2}{2^2}x+c_2$$
$$f^3(x)=frac{3}{2}left(frac{3^2}{2^2}x+c_2right)+c_1=frac{3^3}{2^3}x+c_3$$
and so on.