Show-that-the-minimum-value-of-sinh-x-ncosh-x-is-n-2-1-and-this-occurs-x-0-5ln-n-1-n-1-

Question Number 178354 by Spillover last updated on 15/Oct/22

Show that the minimum value

of \sinh x + ncosh x is \sqrt{n^{2} - 1}

and this occurs x = 0 . 5 \ln (\frac{n - 1}{n + 1})

Answered by Ar Brandon last updated on 16/Oct/22

y = \sinh x + n \cosh x

\frac{d y}{d x} = 0 \Rightarrow \cosh x + n \sinh x = 0

\Rightarrow \coth x = - n \Rightarrow \frac{e^{2 x} + 1}{e^{2 x} - 1} = - n

\Rightarrow (n + 1) e^{2 x} = n - 1 \Rightarrow x_{\min} = \frac{1}{2} \ln (\frac{n - 1}{n + 1})

y_{\min} = \sinh (x_{\min}) + n \cosh (x_{\min})

= \frac{1}{2} (\sqrt{\frac{n - 1}{n + 1}} - \sqrt{\frac{n + 1}{n - 1}}) + \frac{n}{2} (\sqrt{\frac{n - 1}{n + 1}} + \sqrt{\frac{n + 1}{n - 1}})

= \frac{1}{2} (\frac{- 2}{\sqrt{n^{2} - 1}}) + \frac{n}{2} (\frac{2 n}{\sqrt{n^{2} - 1}}) = \frac{n^{2} - 1}{\sqrt{n^{2} - 1}} = \sqrt{n^{2} - 1}

Commented by Spillover last updated on 16/Oct/22

t h a n k s

Commented by Spillover last updated on 17/Oct/22

please explain the second line

from the end .

Commented by Ar Brandon last updated on 17/Oct/22

\sinh (x_{\min}) + n \cosh (x_{\min})

= \frac{1}{2} (e^{x_{\min}} - e^{- x_{\min}}) + \frac{n}{2} (e^{x_{\min}} + e^{- x_{\min}})

= \frac{1}{2} (e^{\frac{1}{2} \ln (\frac{n - 1}{n + 1})} - e^{- \frac{1}{2} \ln (\frac{n - 1}{n + 1})}) + \frac{n}{2} (e^{\frac{1}{2} \ln (\frac{n - 1}{n + 1})} + e^{- \frac{1}{2} \ln (\frac{n - 1}{n + 1})})

e^{\frac{1}{2} \ln (\frac{n - 1}{n + 1})} = e^{\ln {(\frac{n - 1}{n + 1})}^{\frac{1}{2}}} = {(\frac{n - 1}{n + 1})}^{\frac{1}{2}} = \sqrt{\frac{n - 1}{n + 1}}

e^{- \frac{1}{2} \ln (\frac{n - 1}{n + 1})} = \frac{1}{e^{\frac{1}{2} \ln (\frac{n - 1}{n + 1})}}