Is that hyperbolic identity correct?

Question

Given the expression:

\begin{equation} |x|\cosh(kx)+x\sinh(kx), \;\; k>0 \end{equation}

By taking cases for $x$, we have:

\begin{equation} \bullet \quad x>0: x(\cosh(kx)+\sinh(kx))=x\left( \frac{e^{kx}+e^{-kx}}{2}+\frac{e^{kx}-e^{-kx}}{2}\right)=xe^{kx} \\ \bullet \quad x<0: -x(\cosh(kx)-\sinh(kx))=-x\left( \frac{e^{kx}+e^{-kx}}{2}-\frac{e^{kx}-e^{-kx}}{2}\right)=-xe^{-kx} \end{equation}

Is it then possible to claim, due to the above, that:

\begin{equation} |x|\cosh(kx)+x\sinh(kx)=|x|e^{k|x|}, k>0 \end{equation}

holds?

Thank you!

Answer 1

Let's check:

\begin{equation} |x|e^{k|x|} = \begin{cases} x e^{kx} & \text{if } x \geq 0, \\ -x e^{-kx} & \text{if } x < 0. \end{cases} \end{equation}

Yes, it works.