Proof of $f(x)=e^x+x$ injective

Question

I am asked:

Let $f:\mathbb{R} \to \mathbb{R}$ be given by $f(x)=e^x+x$.

Prove that $f$ is injective.

Here's my attempt:

Proof: (Contrapositive) Assume $x_1,x_2\in \mathbb{R}, x_1\neq x_2$. So if $x_1\neq x_2$, then either $x_1<x_2$ or $x_2<x_1$.

If $x_1<x_2$, then $x_2=x_1+a$ for some $a\in \mathbb{R}^+$. Thus, $e^{x_1}+x_1<e^{x_1+a}+(x_1+a)\implies e^{x_1}+x_1<e^{x_1}*e^a+x_1+a\implies \frac{e^{x_1}}{e^{x_1}}<\frac{e^{x_1}*e^a}{e^{x_1}}+\frac{x_1}{e^{x_1}}\implies 1<e^a+\frac{x_1}{e^{x_1}}\implies 0<e^a+\frac{x_1}{e^{x_1}}-1$. Since $a\in \mathbb{R}^+$, $e^a\in \mathbb{R}^+,e^a>1$. Thus, this inequality must be true, so when $x_1<x_2$, $e^{x_1}+x_1<e^{x_2}+x_2\implies f(x_1)\neq f(x_2)$.

Alternatively, if $x_2<x_1$, then $x_1=x_2+b$ for some $b\in \mathbb{R}^+$ Thus, $e^{x_2}+x<e^{x_2+b}+(x_2+b)\implies e^{x_2}+x_2<e^{x_2}*e^b+x_2+b\implies 1<e^b+\frac{x_2}{e^{x_1}}\implies 0<e^b+\frac{x_2}{e^{x_2}}-1$. Since $b\in \mathbb{R}^+$, $e^b\in \mathbb{R}^+,e^b>1$. Thus, this inequality must be true, so when $x_2<x_1$, $e^{x_2}+x_2<e^{x_1}+x_1\implies f(x_1)\neq f(x_2)$.

In either case, $x_1\neq x_2\implies f(x_1)\neq f(x_2)$, which is the contrapositive of the definition of injectivity. Therefore, the function $f$ is injective. $\blacksquare$

Did I miss anything? Thanks!

Answer 1

$$f(x)=e^x+x \implies f'(x)=e^x+1>0$$

Thus the function f(x) is strictly increasing.

Hence, f(x) is one-to-one.

Answer 2

$$e^{x_1}+x_1<e^{x_1}*e^a+x_1+a\implies \frac{e^{x_1}}{e^{x_1}}<\frac{e^{x_1}*e^a}{e^{x_1}}+\frac{x_1}{e^{x_1}}$$ Looks like you made a mistake here, it should be $\frac{a}{e^{x_1}}$ on the right. You make the same mistake with $x_2$ and $b$. However, these mistakes are easily corrected and doesn't change your reasoning. My comment of the direction of "$\Longrightarrow$" I made in my other answer applies here as well.

However, as the others pointed out, this is unnecessarily complicated. $e^x$ is strictly monotone, $x$ is strictly monotone, sums of strictly monotone functions are strictly monotone and strictly monotone functions are always injective.