Continuous Function proof arctangent For some $c\in(-1,1)$ and $x\in[-1,1]$ define $M_c(x)=\frac{x+c}{1+xc}$. Show that $M_c:[-1,1]\rightarrow[-1,1]$ is continuous and bijective with $M_c(\pm1)=\pm1$, $M_c(0)=c$ and inverse $M_c^{-1}=M_{-c}$
I am confused how to prove that, I have the definition of continuity but the definition is given for certain point $x_0$, I don't have clear how to apply the definition in this case. For the bijectivity, I am trying with injectivity and surjectivity.
 A: Hints: To show it is injective, prove that if $M_c(x)=M_c(y)$ then $x=y$ (just solve the equality!). To show it is surjective, show that $t = M_c(x)$ has some solution $x$ if $t\in [-1,1]$.
For continuity, you must show that the relevant equality ($\lim_{x\to x_0} M_c(x) = M_c(x_0)$ for every $x_0\in [-1,1]$.
A: For the continuity part: do you have to prove it using the definition of continuity? I very much hope that you don't — that would be doable, but long and tedious. Instead, recall and use properties of elementary functions. For example, it is an established fact that any rational function $\displaystyle f(x)=\frac{P(x)}{Q(x)}$, where $P(x)$ and $Q(x)$ are polynomials, is continuous on its entire domain, i.e. at all points where the denominator $Q(x)\neq0$. So you should prove that in your example the denominator $Q(x)=1+cx\neq0$ for all $x\in[-1,1]$ and $c\in(-1,1)$, and that will imply the continuity of the function.
For being bijective: yes, show that it is injective and that is it surjective.
