how to determine the lowest number in this curve

Question

can someone help me on how to find the lowest value which the function will get ?

$y=\frac{1}{2}(e^x-e^{-x})+\frac{1}{2}n(e^x+e^{-x})$

thought of using t as $e^x$..but couldn't get an answer.

Answer 1

Method 1. $$2y=e^x(n+1)+e^{-x}(n-1)\implies 2y'=(n+1)e^x -(n-1)e^{-x}$$ setting $y'=0 $ gives $$(n+1)e^{2x}=n-1$$ or $$e^{2x}={n-1\over n+1}$$ so $$e^x=\sqrt {n-1\over n+1}$$ and $$e^{-x}=\sqrt {n+1\over n-1}$$ putting these values in $y={1\over 2}\{e^x(n+1)+e^{-x}(n-1)\}$ we get $$y_{\min}=\sqrt{n^2 -1}$$ which is the same as lab bhattacharjee 's answer in comments (obtained by AM-GM inequality).

Method 2. It is known that if the product of two variables (say $x$,$y$) is a constant, i.e., $xy=c$ then $x+y$ is minimized at points for which $x=y$. Here the product of $(n+1)e^x$ and $(n-1)e^{-x}$ is a constant. Therefore their sum is minimum at points for which $$(n+1)e^x=(n-1)e^{-x}$$ giving the same answer as method 1.