required knowledge for solving this equations Hi everyone there is some kind of functions which they're confusing to me, I already studied whole function knowledge requirements but they wasn't enough for solving new type of questions at all, I studied IB diploma math either. I will write down my questions and I will be so appropriated if you help me about solution.please tell me which book or sources will be helpful for mastering this specific topic of function.


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*Given that $f(x)+f(x+1)=2x+4$, find $f({1\over2})$.


The official solution has suggested using linear function but it doesn't work in other type of same questions (like which I will write below) and it's hard to memorize specific question and it's solution during the test
$$f(x)+f(x+1)=2x+4$$
$$2ax+a+2b=2x+4$$
$$a=1, \quad  b={3\over2}$$
$$f(x)=x+{3\over2}$$
$$f({1\over2})=2$$
I tried to use the same technique for the following question:


*Given that $f(x)+2f({1\over x})=2x+4$, find $f(2)$.


$$ax+b+2\left({a\over x}+b\right)=2x+4$$
$$ax+{2a\over x}+3b=2x+4$$
In question 1 we set $2ax=2x$ so that $a=1$, but that did not work for this question. 
Also I could not solve the following questions: 


*Given that $f(x+y)=f(x)f(y)$ and $f(2)=3$, find $f(4)$.

*Given that $f(xy)=f(x)+f(y)$ and $f(3)=2$, find $f(27)$.
P.S. 
I revisited the first and second question again there is a solution which worked for first one but didn't for second one the official topic says they're both linear function but it doesn't help me solving questions at all by the way:
I said in $f(x)+f(x+1)=2x+4$ we have $f(x+1)$ and that means $f(x)+1$ which moves $f(x)$ vector during $x$ axis so:
$2f(x)+1=2x+4$ and we have $f({1\over 2})$=?
$2f({1\over 2})+1=5$ 
$2f({1\over 2})=4$ 
$f({1\over 2})=2$
but when I tried same solution for second question in this way:
$f(x)+2f({1\over x})=2x+4$
I said $f({1\over x})$ is $x-({x^2-1\over x})$ and for $f(2)$ will be:
$f(2)+2(f(2)-{3\over 2})=4+4$
$3f(2)-3=8$ 
$f(2)= \frac{11}{3}$
this answer is incorrect because the true answer is $f(2) = \frac{2}{3}$
and I didn't understand what is linear function rule in this question 
 A: $f(4)=f(2+2)=f(2)f(2)={f(2)}^2=3^2=9$
for 4th
 $f(27)=f(3.9)=f(3)+f(9=3.3)=f(3)+f(3)+f(3)=3f(3)=3.2=6$
A: The trick here is usually to find good choices of $x$ and $y$ that give us as much information as possible about $f$, and then to use the information we've got and the functional relation to solve the question.
For (3), take $x=y=1$. This gives
$$3 = f(2) = f(1+1) = f(1)^2$$
so that $f(1) = \pm\sqrt{3}$, and $f(3) = f(2+1) = \pm3\sqrt{3}$.
For (4), we have $27=3^3$, so
$$f(27) = f(3\cdot9) = f(3) + f(3\cdot3) = 3f(3) = 6.$$
Notice that taking $x=y=1$ we also have
$$f(1) = 2f(1) \Rightarrow f(1) = 0.$$
A: The functional equation $f(x)+f(x+1)=2x+4$ has many solutions, unless you require additional properties. You can choose $f(\frac{1}{2})$ arbitrary!
Choose, for example, $f(\frac{1}{2})=42$. Then the equation $f(\frac{1}{2})+f(\frac{3}{2})=2\times\frac{1}{2}+4$ implies $f(\frac{3}{2})=5-42=-37$. Further, you have $f(\frac{3}{2})+f(\frac{5}{2})=2\times \frac{3}{2}+4$ which uniquely determines $f(\frac{5}{2})=44$. In this way, you can compute $f(\frac{7}{2})$, $f(\frac{9}{2}), \ldots$, but also $f(-\frac{1}{2})$, $f(-\frac{3}{2})$ etc. For numbers $x$ that are not of the form $\pm\frac{1}{2}$, $\pm\frac{3}{2},\ldots$, you can still keep $f(x)$ to be the original linear function with no change. 
Then $f(x)+f(x+1)=2x+4$ is satisfied for every $x$ and $f(\frac{1}{2})=42$.
A: In the first case, from
$$f(x+1)=-f(x)+2x+4,$$
we deduce 
$$f(x+2)=-f(x+1)+2(x+1)+4=f(x)-2x-4+2x+2+4=f(x)+2$$ and by recurrence,
$$f(x+2k)=f(x)+2k.$$
We can freely choose $f(x)=\phi(x)$ for $x$ in the range $[0,1)$, and this extends $f$ to all reals by
$$f(x+2k)=\phi(x)+2k,\\
f(x+2k+1)=-\phi(x)+2x+2k+4.$$
In particular, we can form continuous solutions with $\phi$ continuous and $\phi(1)=6-\phi(0)$.
