Is it always possible to transform one function into another? Suppose f(x) = g(x) for all real x, and both f and g are sufficiently nice (perhaps we might limit them to be polynomials or analytic functions). Can we always manipulate one (with algebraic transformations) into the other? Is the same true for functions with multiple arguments?
To say more precisely what I mean let me quote Arthur's comment:

interested in the expressions of functions, like having the two functions $f(x)=\frac{4x+2}{2}$ and $g(x)=2x+1$ and use algebra to transform one into the other

 A: That depends on what kind of restrictions you put on the expressions defining $f$ and $g$, and what your exact definition of algebraic transformations is.
For example, $$
  \lim_{n\to\infty} \left(1 + \frac{x}{n}\right)^n = e^x = \lim_{n\to\infty}\sum_{i=0}^n \frac{x^i}{i!}
$$
yet simply applications of algbraic identities won't prove the equivalence, since $$
   \left(1 + \frac{x}{n}\right)^n \neq \sum_{i=0}^n\frac{x^i}{i!} \text{.}
$$
On the other hand, for polynomials algebraic identifies are sufficient to prove or disprove $p(x) = q(x)$ - it's sufficient to fully expand both sides, and compare the coefficients.
If you restrict yourself to functions definable in the first-order language of real closed fields, and $\forall x\, f(x) = g(x)$ is true over $\mathbb{R}$, then you can prove that by using just the axioms for RCF (real close fields) and the rules of deduction of first-order logic. In other words, if $$
  \mathbb{R} \models \forall x\, f(x) = g(x)
$$
then $$
  \textrm{RCF} \vdash \forall x\, f(x) = g(x) \text{.}
$$
This works because RCF is complete, i.e. for every possible logical statement $\varphi$, RCF either proves $\varphi$ or its negation $\lnot \varphi$.
A: I am assuming you are asking if this equality can be extended to the entire complex plane.  The answer for polynomials and analytic functions is yes.
Polynomials of degree $n$ are completely determined by $n+1$ points.  So if two polynomials agree on the real line, they must be the same polynomial.
This same idea can be extended to analytic functions.  If the zeros of an analytic function have an accumulation point inside of its domain, then the function must be the zero function.  Since, $f-g$ is an analytic function (if $f$ and $g$ are) then $f(x)-g(x)=0$ for all real $x$.  Thus the function $f-g$ has an accumulation of zeros at every real number.  Therefore, $f-g=0$ on the complex plane, and $f=g$.
