Determine the x-intercept in terms of n

[Saracen Rue]

Homework Statement

Determine the exact value of the x-intercept in terms of n, where n>0.

Relevant Equations

I presume just some rearranging of equations.

Here's what the picture in the textbook looks like (it kept being too unclear when trying to take a picture so I had to replicate it on Desmos, however my replication is identical to the graph provided in the question.

As for the rest of the question... I'm completely stumped. I should note that this is a challenge question, but even still I can't work out anyway to solve for an exact answer for this because the only way for the funtion to equal 0 would be for arccoth(nx+2) to equal 0, however coth(0) isn't definied so there's no way to continue.

I know that the function definitely does have an x-intercept for every value on n>0 because when graphing it using Desmos and having n as a slider it shows that there is always 1 x-intercept.

However, even when I let n=1 and try to solve (for x when the function is 0) using Wolfram Alpha it says no solutions exist (see below image)

I'm really stuck with this one and I'm not sure where to go from here.

Any help is greatly appreciated. Thanks for your time :)

 

[scottdave]

Perhaps think about what value of the Arcoth() argument would make it equation zero? http://mathworld.wolfram.com/InverseHyperbolicCotangent.html

 

[Ray Vickson]

View attachment 241406
Here's what the picture in the textbook looks like (it kept being too unclear when trying to take a picture so I had to replicate it on Desmos, however my replication is identical to the graph provided in the question.

As for the rest of the question... I'm completely stumped. I should note that this is a challenge question, but even still I can't work out anyway to solve for an exact answer for this because the only way for the funtion to equal 0 would be for arccoth(nx+2) to equal 0, however coth(0) isn't definied so there's no way to continue.

I know that the function definitely does have an x-intercept for every value on n>0 because when graphing it using Desmos and having n as a slider it shows that there is always 1 x-intercept.

However, even when I let n=1 and try to solve (for x when the function is 0) using Wolfram Alpha it says no solutions exist (see below image) View attachment 241407
I'm really stuck with this one and I'm not sure where to go from here.

Any help is greatly appreciated. Thanks for your time :)

If $nx-1<0$ then $(+\infty)^{nx-1} = 0.$

 

[Saracen Rue]

Perhaps think about what value of the Arcoth() argument would make it equation zero? http://mathworld.wolfram.com/InverseHyperbolicCotangent.html

I tried that, but when arccoth(nx+2)=0, nx+2=coth(0)=∞. This doesn't help with being able to solve for x in terms of n.

If $nx-1<0$ then $(+\infty)^{nx-1} = 0.$

But to solve that for x you need to say $log_{\infty}(0)=nx-1$, and $log_{\infty}(0)$ isn't defined so this doesn't help find x in terms of n.

 

[scottdave]

I don't think that you need to take the log. Also, to be clear is the argument of Arcoth raised to the power? Or do you take Arcoth first then raise to the power?

 

[Ray Vickson]

I tried that, but when arccoth(nx+2)=0, nx+2=coth(0)=∞. This doesn't help with being able to solve for x in terms of n.But to solve that for x you need to say $log_{\infty}(0)=nx-1$, and $log_{\infty}(0)$ isn't defined so this doesn't help find x in terms of n.

Of course not; you need to take limits. The symbol $\infty$ is just a shorthand notation for a limit.

 

[Saracen Rue]

I don't think that you need to take the log. Also, to be clear is the argument of Arcoth raised to the power? Or do you take Arcoth first then raise to the power?

Sorry for the poor notation; it should be $(arccoth(nx+2))^{(nx-1)}$

Of course not; you need to take limits. The symbol $\infty$ is just a shorthand notation for a limit.

Right, I'm still a bit confused as to how I could use limits to create a general relationship between x and n.

 

[SammyS]

Sorry for the poor notation; it should be $(\text{arccoth}(nx+2))^{(nx-1)}$

Right, I'm still a bit confused as to how I could use limits to create a general relationship between x and n.

That clarification is helpful.

One idea is to let $u=nx$.

Graph $(\text{arccoth}(u+2))^{(u-1)}$ versus $u$.

 

[Saracen Rue]

That clarification is helpful.

One idea is to let $u=nx$.

Graph $(\text{arccoth}(u+2))^{(u-1)}$ versus $u$.

Hi, thank you for your help. After playing with the function more and I think I've worked out how to determine the intercept.