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

Thanks!

Thanks!

- Thread starter Eric78
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- #1

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

Thanks!

Thanks!

- #2

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You should be able to reduce everything to a fourth order algebraic equation with integer coefficents.Eric78 said:Hi, I need to find the Y-coordinate of the extrema of cosh(x)+tanh(x),

Thanks!

The condition for extrema (giving the critical points of the graphic) is

[tex] (\cosh x+\tanh x)'=0 [/tex] which gives an transcedental equation in the hyperbolic functions [tex] \cosh x [/tex] and [tex] \sinh x [/tex].Substitute the definiton of these functions and then make the substitution [tex] \exp x = \lambda [/tex].The equation for lambda is as i said and look for its positive roots.

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Tom Mattson

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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 [Broken]

1. The rational root test

and if no rational roots then...

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

Last edited by a moderator:

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Zurtex

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[tex]y = \cosh x + \tanh x[/tex]

[tex]y \cosh x = \cosh^2 x + \sinh x[/tex]

[tex]y' \cosh x + y \sinh x = 2 \cosh x \sinh x + \cosh \x[/tex]

[tex]y'=0[/tex]

[tex]\cosh^2 x + \sinh x = 2 \cosh x \sinh \x + \cosh \x[/tex]

[tex]\cosh^2 x + (\sinh x - \cosh x) - 2 \cosh x \sinh x = 0[/tex]

[tex]\frac{e^{2x} + 2e^x - 2e^{-2x}}{4} + e^{-x} - \frac{e^{2x} - e^{-2x}}{2} = 0[/tex]

[tex]e^{2x} + 2e^x - e^{-2x} + 4e^{-x} - 2e^{2x} + 2e^{-2x}=0[/tex]

[tex]-e^{2x} + 2e^x + 4e^{-x} + e^{-2x}=0[/tex]

[tex]-e^{4x} + 2e^{3x} + 4e^x + 1=0[/tex]

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

- #6

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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.Eric78 said:

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