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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 $a=b$.

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  • $\begingroup$ "but isn't this like solving the equation x=ex" don't need to solve it. Just need to know a solution exists. $\endgroup$
    – fleablood
    Jul 31 at 5:43
  • $\begingroup$ Also. If $a < b$ then $e^a < e^b$ and if $a \ge 0$ then $ae^a < be^b$ (will have to do something for $x < 0$.) $\endgroup$
    – fleablood
    Jul 31 at 5:46
  • $\begingroup$ Agreed to that; "just need to show a solution exist" - thanks! $\endgroup$
    – NumberMan
    Jul 31 at 5:49
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$\frac d {dx} (xe^{x})=xe^{x}+e^{x}=(1+x)e^{x} >0$ for $x >-1$. So $xe^{x}$ is strictly increasing on $(-1,\infty)$. Can you finish?

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    $\begingroup$ Yes... that was easy! I got so stuck in the method I tried in the question. Thanks! $\endgroup$
    – NumberMan
    Jul 31 at 5:35
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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.

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