Whether the function is surjective or not? How to check whether a function is surjective or not?
  function is $$f(x)=(x^{2}+1)^{35}$$ where $x\in \mathbb{R}$
   How to write $F(x)$ in terms of $y$?
 A: If you consider it as a function $f:\Bbb R\to\Bbb R$, or even $f:\Bbb R\to[0,\infty)$, then it is not surjective. Indeed, $(x^2+1)^{35}\geq (0+1)^{35}=1^{35}=1$, so the range of $f$ is contained in $[1,\infty)$. 
Actually, the range is precisely $[1,\infty)$, so the function
$$f:\Bbb R\to [1,\infty),\quad f(x)=(x^2+1)^{35}$$
is surjective.
A: In general if $f:X\to Y$ is a function to be checked on surjectivity then find out the range of the function (i.e. the set $\{f(x)\mid x\in X\}\subseteq Y$). The function is surjective if and only if its range coincides with $Y$.
Alternatively you can wonder: can I find a function $g:Y\to X$ such that $f\circ g: Y\to Y$ is the identity on $Y$? The answer will be "yes" if and only if $f$ is surjective.
Your specific question can only be answered if $Y$ is known, wich is not the case.
A: For this particular case, you can exhibit numbers which are not included in the image of $f$, 0 for instance :
$\forall x\in \mathbb R, (x^2+1)^{35}\geqslant 1$
so any $x<1$ is not attained by f. Hence f is not surjective.
