Different type of function equation

Question

I've spent three days revising functions, although it wasn't enough at all because I think my sources weren't good enough to allow me to become a master in the field of math. By the way, there are two questions which remain because I couldn't solve them.

1. $$f:N^2>>>R$$ $$f(n):\left({1\over n}\right)*f(n+1)$$ $$f(1)=1$$ $$f(7)=?$$

I tried this solution, but I didn't understand the $f(1)=1$ rule

$f(n)=\left({fg(n)\over n}\right)$

$\left({fg(n)\over n}\right)=f(n+1)$

$f(n)*n=fg(n)$

$g(n+1)=n$

$n+1=17$

$n=16$

The answer keys says correct answer is 16! but I think ! is typing mistake,isn't it?

2. $$f(x)=\left({2x*f(x-1)\over x+1}\right)$$ $$(1)=1$$ $$f(7)=?$$ the way I tried to solve was:

$fg(x)=f(x-1)$

$f(7)=x-1$ and $x=8$

Then I tried lots of methods. They were all wrong.

Tonight I want to finish this subject. It took me lots of extra hours and kept me behind my schedule. I asked many questions tonight, thank you everyone for your help. My concern is for which method I should use to solve this questions in a minute during my exam. I will write my other examples below to show the type of questions which I had been solving the problem with (they're solved no need to resolve) :x

1. $f(x)+f(x+1)=2x+4$ find $f\left({1\over2}\right)=\text{?}$

2. $f(x)+2f\left({1\over x}\right)=2x+4$ find $f(2)=\text{?}$

3. $f(x)=3x-5$ find $f(2x+1)=?$

4. $f(-x)=2f(x)+6$ find $f(3)=?$

Would you tell me what my weakness in this topic (functions) is, and how I can conclude the topic effectively?

Answer 1

Are you sure the answer key did not say $$f(7) = 6!$$

The way to solve this is to start by manipulating the equation to get $$ f(n+1) = n f(n) $$ from which it becomes clear that $f(1) = 1; f(2) = 1\cdot 1 = 1; f(3) = 2 \cdot 1 = 2; f(4) = 3 \cdot 2 = 6 \ldots$ and in general $$ f(k) = (k-1)! $$