Show that the minimum value of sinh x+ncosh x is (√(n^2 −1)) and this occurs x=0.5ln (((n−1)/(n+1)))


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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

	$\sinh (x_{\min}) + n \cosh (x_{\min})$ $= \frac{1}{2} (e^{x_{\min}} - e^{- x_{\min}}) + \frac{n}{2} (e^{x_{\min}} + e^{- x_{\min}})$ $= \frac{1}{2} (e^{\frac{1}{2} \ln (\frac{n - 1}{n + 1})} - e^{- \frac{1}{2} \ln (\frac{n - 1}{n + 1})}) + \frac{n}{2} (e^{\frac{1}{2} \ln (\frac{n - 1}{n + 1})} + e^{- \frac{1}{2} \ln (\frac{n - 1}{n + 1})})$ $e^{\frac{1}{2} \ln (\frac{n - 1}{n + 1})} = e^{\ln {(\frac{n - 1}{n + 1})}^{\frac{1}{2}}} = {(\frac{n - 1}{n + 1})}^{\frac{1}{2}} = \sqrt{\frac{n - 1}{n + 1}}$ $e^{- \frac{1}{2} \ln (\frac{n - 1}{n + 1})} = \frac{1}{e^{\frac{1}{2} \ln (\frac{n - 1}{n + 1})}}$
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