I've just started learning functions but my algebra is letting me down:
If $f(x)$ = $\frac{1}{x+1}$, prove that $f^2(x)$ = $\frac{x+1}{x+2}$. Hence prove $f^3(x) ...$
Am I right in thinking that $f^2(x)$ = $\frac{1}{\frac{1}{x+1}+1}$?
After working through, I end up with $\frac{x+1}{1}$
On
As I explained in my previous answer, following that method gives us: $$f^3(x)=f(f(f(x))) = f(\frac{x+1}{x+2}) = \frac{x+2}{2x+3}$$
So, yes, you have correctly computed $f^2(x)$ but not $f^3(x)$.
On
$$\frac{1}{\frac{1}{x+1}+1} = \frac{1}{\frac{x+2}{x+1}} = \frac{x+1}{x+2}$$ so your approach is right but you made a small mistake while manipulating the expression I believe.
On
$$f(x)=\frac1{x+1},$$ $$f^2(x)=\frac1{\frac1{x+1}+1}=\frac{x+1}{x+2},$$ $$f^3(x)=\frac1{\frac{x+1}{x+2}+1}=\frac{x+2}{2x+3},$$ $$f^4(x)=\frac1{\frac{x+2}{2x+3}+1}=\frac{2x+3}{3x+5},$$
$$f^n(x)=\frac{F_{n-1}x+F_n}{F_nx+F_{n+1}},$$
where $F_n$ is the $n^{\rm th}$ Fibonacci number.
Proof by induction: for $n=1$, $$f(x)=\frac1{x+1}=\frac{F_0x+F_1}{F_1x+F_2},$$
If the formula is true for $n=k$, then $$f^k(x)=\frac{F_{k-1}x+F_k}{F_kx+F_{k+1}},$$
and $$f^{k+1}(x)=\frac1{f^k(x)+1}=\frac1{\frac{F_{k-1}x+F_k}{F_kx+F_{k+1}}+1}=\frac{F_kx+F_{k+1}}{F_{k-1}x+F_k+F_kx+F_{k+1}}=\frac{F_kx+F_{k+1}}{F_{k+1}x+F_{k+2}}.$$
QED.
You are correct, here is $f^2(x)$ as a hint$$f^2(x) = \frac{1}{\frac{1}{x+1}+1} = \frac{1}{\frac{x+2}{x+1}} = \frac{x+1}{x+2}$$