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**geton** Given that y = sinh x + k cosh x, show that the least value of y is √(k^2 - 1) and that this occurs at x = ½ ln (k-1)/(k+1) where k is a constant and |k| > 1.

What is mean by the least value?

__My solution__:

$\displaystyle

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

$\displaystyle = \frac {1}{2}(e^x - e^{-x} + ke^x + ke^{-x})$

$\displaystyle = \frac {1}{2} \left(e^x (1+k) - e^{-x} (1-k)\right)$

$\displaystyle = \frac {1}{2} \left( \left(\frac {k-1}{k+1}\right)^\frac{1}{2} (k+1) + \left(-\frac{k-1}{k+1}\right)^\frac{1}{2} (k-1)\right)$ [Substitute x = ½ ln (k-1)/(k+1) ]

By simplifying this I got $\displaystyle \left(\frac {k-1}{k+1}\right)^\frac {1}{2}$

Is that a right way? Mr F says: No.

Now what to do?