# Math Help - Hyperbolic function

1. ## Hyperbolic function

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:

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

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

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

Is that a right way? Now what to do?

2. Originally Posted by 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:

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

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

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

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

Now what to do?
1. Show that y has a single stationary point by finding the solution single solution to dy/dx = 0.

2. Show that this stationary point is a minimum turning point by testing its nature.

3. You have the x-coordinate of the turning point, now find the y-coordinate.

3. Yes, I solved that. Thanks.