Showing that $f(x)=2xe^{x}+1$ is injective for $x \geq -1$

Question

How do you show that $f(x)=2xe^{x}+1$ is injective for $x \geq -1$?

If you choose two points $a,b \in D_f$, and say $f(a)=f(b)$ you'll get $ae^a=be^b$, but isn't this like solving the equation $x=e^x$? The idea was to end up with $a=b$.

Answer 1

$\frac d {dx} (xe^{x})=xe^{x}+e^{x}=(1+x)e^{x} >0$ for $x >-1$. So $xe^{x}$ is strictly increasing on $(-1,\infty)$. Can you finish?

Answer 2

Showing that the function is strictly increasing is enough to show that it's injective. (In general, strictly monotonic functions are injective)

Simply take the derivative and check that it's strictly greater than 0.