What kinds of functions have fixed points? Among continuous functions, can we characterize those which have fixed points and those which do not?
Geometrically, these are the functions that intersect the line $f(x) = x$. Is that the most concise description we can give?
For instance, all polynomials of degree $0$ (constant functions $f(x) = c$) have fixed points, not all polynomials of degree $1$ or $2$ do, but all cubic functions do (since they start with one sign and end with the other, they must cross the line). In general all odd-degree polynomials have at least one fixed point.
But some quadratics (e.g. $f(x) = x^2 - 1$) have fixed points, while others ($x^2 + 1$) do not. What's the difference?
 A: I don't know of a truly general result, but one result is that every contraction map from a complete metric space $(X,d)$ to itself, i.e., a map $f$ so that $d(x,y)>d(f(x),f(y))$ has a (unique) fixed point, and there is a formula for it.
There are many other fixed-point results, like Brouwer's fixed point theorem that says that every continuous map from the closed unit sphere in $\mathbb R^n$ to itself must have a fixed point. 
There are many other results, including the Banach Fixed Point thm, but no one overarching one that I am familiar with:
https://en.wikipedia.org/wiki/Fixed-point_theorem
A: Fixed point of a function f(x) are those $x\in \mathbb{R}$ such that $f(x)=x$.
For the case $f(x)=x^2+1$, the fixed points of $f(x)$ are $x\in \mathbb{R}$ such that $x^2+1=x$. So arranging this gives $x^2-x+1=0$, with a=1, b=-1 and c=1 when compared with $ax^2+bx+c=0$. Now, $b^2-4ac=1-4=-3$. So $\sqrt{b^2-4ac}=\sqrt{-3}$ does not have a solution in the set of real numbers. Thus, $f(x)=x^2+1$ has no fixed point in $\mathbb{R}$. On the other hand, $f(x)=x^2-1$ (by a similar computation as above) has fixed points in $\mathbb{R}$ since $\sqrt{b^2-4ac} \geq 0$.
