Let f and g be two real functions...

Let f and g be two real functions...

Postby Learningforcomp » Mon Sep 21, 2020 9:24 am

Let f and g be two real functions such that g is the inverse of f is defined as
bt.png
bt.png (1.51 KiB) Viewed 388 times
, calculate
nu.png
nu.png (911 Bytes) Viewed 388 times
and select the correct option:

kerute.png
kerute.png (4.2 KiB) Viewed 388 times


how do I solve this?
Your help is deeply appreciated.
Thank you.
Learningforcomp
 
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Re: Let f and g be two real functions...

Postby Baltuilhe » Mon Sep 21, 2020 2:23 pm

Good afternoon!

[tex]y=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}[/tex]

[tex]y=\frac{e^{x}-\frac{1}{e^{x}}}{e^{x}+\frac{1}{e^{x}}}[/tex]

[tex]y=\frac{\frac{e^{2x}-1}{e^{x}}}{\frac{e^{2x}+1}{e^{x}}}[/tex]

[tex]y=\frac{e^{2x}-1}{e^{2x}+1}[/tex]

[tex]y\cdot\left(e^{2x}+1\right)=e^{2x}-1[/tex]

[tex]y\cdot e^{2x}-e^{2x}=-1-y[/tex]

[tex]e^{2x}\cdot\left(y-1\right)=-\left(1+y\right)[/tex]

[tex]e^{2x}=\frac{y+1}{y-1}[/tex]

[tex]2x=\ln\left(\frac{y+1}{y-1}\right)[/tex]

[tex]x=g(y)=\frac{1}{2}\cdot\ln\left(\frac{y+1}{y-1}\right)[/tex]

Try to reach the answer, now :)

Good luck :)

Baltuilhe
 
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Re: Let f and g be two real functions...

Postby nathi123 » Tue Sep 22, 2020 5:20 am

[tex]e^{2x}(y-1)=-1-y\Leftrightarrow e^{2x}=\frac{-1-y}{y-1}\Leftrightarrow e^{2x}=\frac{1+y}{1-y}\Rightarrow lne^{2x}=ln\frac{1+y}{1-y}\Leftrightarrow 2x=ln\frac{1+y}{1-y}[/tex]
[tex]x=\frac{1}{2}ln\frac{1+y}{1-y}=ln\sqrt{\frac{1+y}{1-y}}=g(y)\Rightarrow g(\frac{1}{2})=ln\sqrt{\frac{1+\frac{1}{2}}{1-\frac{1}{2}}}=ln\sqrt{3}\Rightarrow e^{g(\frac{1}{2})}=\sqrt{3}[/tex]

nathi123
 
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Re: Let f and g be two real functions...

Postby Guest » Tue Sep 22, 2020 12:24 pm

Baltuilhe wrote:Good afternoon!

[tex]y=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}[/tex]

[tex]y=\frac{e^{x}-\frac{1}{e^{x}}}{e^{x}+\frac{1}{e^{x}}}[/tex]

[tex]y=\frac{\frac{e^{2x}-1}{e^{x}}}{\frac{e^{2x}+1}{e^{x}}}[/tex]

[tex]y=\frac{e^{2x}-1}{e^{2x}+1}[/tex]

[tex]y\cdot\left(e^{2x}+1\right)=e^{2x}-1[/tex]

[tex]y\cdot e^{2x}-e^{2x}=-1-y[/tex]

[tex]e^{2x}\cdot\left(y-1\right)=-\left(1+y\right)[/tex]

[tex]e^{2x}=\frac{y+1}{y-1}[/tex]

[tex]2x=\ln\left(\frac{y+1}{y-1}\right)[/tex]

[tex]x=g(y)=\frac{1}{2}\cdot\ln\left(\frac{y+1}{y-1}\right)[/tex]

Try to reach the answer, now :)

Good luck :)

Hello, and thank you for your answer.
I have a question, You wrote that: e^2x⋅(y−1)=−(1+y) and then e^2x=y−1/y+1​, where did the subtraction sign of -(1+y) go?
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Re: Let f and g be two real functions...

Postby Guest » Fri Nov 06, 2020 4:44 pm

That must have been a typo. [tex]\frac{-(y+1)}{y-1}= \frac{y+1}{-(y-1)}= \frac{y+1}{1- y}[/tex].
Guest
 

Re: Let f and g be two real functions...

Postby Guest » Sat Nov 21, 2020 12:26 pm

If g(x) is the inverse function to [tex]f(x)= \frac{e^x- e^{-x}}{e^x+ e^{-x}}[/tex] then g(1/2) is the value of x such that [tex]f(x)= \frac{e^x- e^{-x}}{e^x+ e^{-x}}= \frac{1}{2}[/tex].

So [tex]2(e^x- e^{-x})= e^x+ e^{-x}[/tex]
[tex]2e^x- 2e^{-x}= e^x+ e^{-x}[/tex]

[tex]e^{x}= 3e^{-x}[/tex]
[tex]e^{2x}= 3[/tex]
[tex]2x= ln(3)[/tex]
[tex]x= \frac{ln(x)}{2}[/tex].
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