Prove that the function $\cosh: [0, \infty[ \rightarrow \mathbb{R}$ is injective
I have tried to prove $f(x_1) = f(x_2) \implies x_1=x_2$
I can not figure out how to prove $\frac{e^{x_1} + e^{-x_1}}2 = \frac{e^{x_2} + e^{-x_2}}2 \implies x_1=x_2 \forall x_1,x_2 \in [0, \infty[ $
Put $x = e^{x_1}, y = e^{x_2} \implies x+\dfrac{1}{x} = y + \dfrac{1}{y} \implies \dfrac{(x-y)(xy-1)}{xy} = 0 \implies x - y = 0$ or $xy = 1$ since $x, y \ge 1$. If $x - y = 0 \implies x = y \implies x_1 = x_2$. If $xy = 1 \implies x = y = 1 \implies x_1 = x_2 = 0 \implies x_1 = x_2$. Thus either case yields a $1-1$ function for $\cosh$