What is the Y-coordinate of the extrema for cosh(x)+tanh(x)?

[Eric78]

Hi, I need to find the Y-coordinate of the extrema of cosh(x)+tanh(x),

Thanks!

 

[dextercioby]

Hi, I need to find the Y-coordinate of the extrema of cosh(x)+tanh(x),

Thanks!

You should be able to reduce everything to a fourth order algebraic equation with integer coefficents.
The condition for extrema (giving the critical points of the graphic) is
$$(\cosh x+\tanh x)'=0$$ which gives an transcedental equation in the hyperbolic functions $$\cosh x$$ and $$\sinh x$$.Substitute the definition of these functions and then make the substitution $$\exp x = \lambda$$.The equation for lambda is as i said and look for its positive roots.

 

[Eric78]

Yes, I found x^3+x+1=0 where x=sinh(x), but there is no simple root, and I need a quite simple solution...

 

[quantumdude]

I'd hunt the root down in 2 steps:

1. The rational root test

and if no rational roots then...

2. http://www.math.sc.edu/cgi-bin/sumcgi/Newton.pl

 

[Zurtex]

Here is my attempt at the problem:

$$y = \cosh x + \tanh x$$

$$y \cosh x = \cosh^2 x + \sinh x$$

$$y' \cosh x + y \sinh x = 2 \cosh x \sinh x + \cosh \x$$

$$y'=0$$

$$\cosh^2 x + \sinh x = 2 \cosh x \sinh \x + \cosh \x$$

$$\cosh^2 x + (\sinh x - \cosh x) - 2 \cosh x \sinh x = 0$$

$$\frac{e^{2x} + 2e^x - 2e^{-2x}}{4} + e^{-x} - \frac{e^{2x} - e^{-2x}}{2} = 0$$

$$e^{2x} + 2e^x - e^{-2x} + 4e^{-x} - 2e^{2x} + 2e^{-2x}=0$$

$$-e^{2x} + 2e^x + 4e^{-x} + e^{-2x}=0$$

$$-e^{4x} + 2e^{3x} + 4e^x + 1=0$$

Have I made a mistake or something? Looks pretty hard to me (although solvable exactly).

 

[dextercioby]

Yes, I found x^3+x+1=0 where x=sinh(x), but there is no simple root, and I need a quite simple solution...

You're right.This is the quations u should be getting.There is at least one solution in R(which should be negative and in the interval (-1,0)),but to find all of them u have to apply Cardano's formulas.My guess is the other two are complex.And sinh(x) can take only real values,so the path to finding the extremas is hopefully clear from now.