what is the relation between $f(x+1)$ and $f(x)$? I searched so much over math sites and google but I didn't find helpful hints and required knowledge or the specific name of this topic in function.
I stuck in relation and operations on $f(x)$ which is really important for solving problems.
for example  relation between $f(x)$ and $f(x+1)$
in more complicated form for instance:
1. $$f(x)=3x-5$$
$$f(2x+1)=?$$
true answer is $$2f(x)+8$$
I tried this solution but it doesn't work
$f(2)$*$f(x)$+$f(1)$=?
$f(1)$=$-2$
$f(2)$=$2*3-5=1$
$f(2)$*$f(x)$+$f(1)$=?
$1(3x-5)-2$=$3x-7$ which gives us $f(x)-2$
another instance:
$$f(-x)=2f(x)+6$$
$$f(3)=?$$
true answer is $$-6$$
I tried this solution 
$f(x)(-1)$=$2f(x)+6$
$-f(x)$=$2f(x)+6$
$-3f(x)$=$6$
$f(x)$=$-2$
so I will be really appreciative if you tell me the relation between operation on $f(x)$ like $f({1\over x})$ , $f(x-1)$  and ..

Hint: For $f(x)=3x-5$.  The expression $f(2x+1)$ means that you substitute $2x+1$ every time you saw $x$ in the original expression.  Therefore,
\begin{align*}
f(2x+1)&=3(2x+1)-5\\
&=6x+3-5\\
&=6x-2.
\end{align*}
Since the goal is to write this in terms of $f(x)$, we try to put this expression into the form for $f(x)$, namely we see the $3x$ term in $f(x)$ while $f(2x+1)$ has a term of $6x$, which is twice $3x$.  So, we expect the form to be $2f(x)=6x-10$.  The problem is that the constant term doesn't line up correctly.  To correct the constant term, we must add and subtract $8$ to get
$$
f(2x+1)=6x-2=6x-2-8+8=6x-10+8=2f(x)+8.
$$
