Mathematics Inclined Week #1: Medium Category

Hello, I am Sxerks, an avid Brainly user. I'm not sure if this has been done, nor do I think I'll keep this up to date, but I'm interested to see how this goes. For those who are looking for quick points, I will be posting up a range of easy to difficult problems each week.

The range of points are dependent of the difficulty of each problem, in my eyes. My solutions will be posted a week later, and a new problem will be posted. Since only two answers are allowed, it's first come, first serve. Those who provide no working out will have theirs removed, hopefully, so someone else can chip in. Discussion is allowed.

Week 1: Medium Category.

Problem: Find the inverse of:
[tex]y = ln(\frac{x - \sqrt{x^{2} - 4}}{2})[/tex]

Question

06-04-2017
Mathematics

contestada

Mathematics Inclined Week #1: Medium Category

Hello, I am Sxerks, an avid Brainly user. I'm not sure if this has been done, nor do I think I'll keep this up to date, but I'm interested to see how this goes. For those who are looking for quick points, I will be posting up a range of easy to difficult problems each week.

The range of points are dependent of the difficulty of each problem, in my eyes. My solutions will be posted a week later, and a new problem will be posted. Since only two answers are allowed, it's first come, first serve. Those who provide no working out will have theirs removed, hopefully, so someone else can chip in. Discussion is allowed.

Week 1: Medium Category.

Problem: Find the inverse of:
[tex]y = ln(\frac{x - \sqrt{x^{2} - 4}}{2})[/tex]

Respuesta :

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LammettHash LammettHash · Answer 1 · 2017-04-07T22:47:41+02:00

[tex]y=\ln\dfrac{x-\sqrt{x^2-4}}2[/tex]
[tex]\implies e^y=\dfrac{x-\sqrt{x^2-4}}2[/tex]
[tex]\implies e^{-y}=\dfrac2{x-\sqrt{x^2-4}}[/tex]

[tex]\implies e^y+e^{-y}=\dfrac{x-\sqrt{x^2-4}}2+\dfrac2{x-\sqrt{x^2-4}}[/tex]
[tex]e^y+e^{-y}=\dfrac{(x-\sqrt{x^2-4})^2+4}{2(x-\sqrt{x^2-4})}[/tex]
[tex]e^y+e^{-y}=\dfrac{2x(x-\sqrt{x^2-4})}{2(x-\sqrt{x^2-4})}[/tex]
[tex]e^y+e^{-y}=x[/tex]

[tex]\implies y=e^x+e^{-x}=2\cosh x[/tex] is the inverse, though this isn't exactly right because [tex]2\cosh x[/tex] is not invertible over its entire domain. So we need to restrict it appropriately. We have [tex]\ln\dfrac{x-\sqrt{x^2-4}}2\le0[/tex] for all [tex]x\ge2[/tex], which means we desire [tex]2\cosh x\ge2[/tex] for all [tex]x\le0[/tex], and so the restricted domain needs to be all non-positive real numbers.