is the function Bijective?

[Randyyy]

[MATH]f(x) = x \mid x \mid +1[/MATH], is the following function bijective, if so, what´s the inverse? Note, you must also state the domain and range of the inverse and the function.
\begin{equation*}
\lvert x \rvert = \left\{
\begin{array}{ll}
-x & \quad x \leq 0 \\
x & \quad x \geq 0
\end{array}
\right.
\end{equation*}
[MATH]g(x)=x^2+1, x \geq 0 [/MATH][MATH]h(x)=-x^2+1, x \leq 0[/MATH]
[MATH]h^-1(x)=\pm \sqrt{x-1}, x\geq 1[/MATH][MATH]g^-1(x)=\pm \sqrt{1-x}, x\leq 1[/MATH]
Is this correct so far or am I going in the wrong direction? I have a hard time knowing which sign I should choose for my inverses, are there any "tricks" to knowing?

 

[Randyyy]

Sitting with it a bit I figured and have come this far.
I noticed I made a mistake, Abs(x), x=> 0 or -x < 0, so I correct that and also figured out the sign for my inverses.

[MATH]g(x)=x^2+1, x\geq 0[/MATH][MATH]h(x)=-x^2+1, x < 0[/MATH][MATH]g^{-1}(x)=\sqrt{x-1}[/MATH], [MATH]x \geq 1[/MATH][MATH]h^{-1}(x)=-\sqrt{1-x}[/MATH], [MATH]x \leq 1[/MATH]Now I just have fuse them as one with a piecewise.

 

[Dr.Peterson]

I think that looks good, if your intention is that g and h are the two "pieces" of f, and [MATH]g^{-1}[/MATH] and [MATH]h^{-1}[/MATH] the pieces of [MATH]f^{-1}[/MATH].

You can check it with a graph:

 

[Randyyy]

Yes, that is what I meant. So it seems that f(x) (the original function) has got a domain and codomain of R if I fuse g(x) and h(x) into f(x) and the same with the inverses.

 

[Dr.Peterson]

Yes. The domain of the piecewise function is the union of the domains of the pieces.

 

[Randyyy]

Thank you for the help!