Given that #cosh3x+3coshx=4cosh^3x# how do you show that #k^3 = 3k - 4# has a root in between #-3# and #-2# and choose a suitable value of #y# for #k=ycoshx# and calculate the root?

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Answer 1 · 2016-12-14T15:55:19

The real root is #k = -2.195823345445647#

Making #k = y cosh(x)# and substituting into #k^3 = 3k - 4# we have

#y^3cosh^3(x)=3ycosh(x)-4#

dividing both sides by #y ne 0# we have

#y^2cosh^3(x)x=3cosh(x)-4/y#

making now #y = -2# and substituting we have

#4cosh^3(x)=3cosh(x)+2#

comparing now with the identity

#cosh(3x)+3cosh(x)=4cosh^3(x)#

we conclude

#cosh(3x)=2#

Solving for #x# we get #x = pm0.4389859656416055#

and finally

#k = -2cosh(x) = -2.195823345445647# which is the sougth real root.

As we can observe

#-3 < -2.195823345445647 < -2#