Let f and g be two real functions...

[Learningforcomp]

Let f and g be two real functions such that g is the inverse of f is defined as

bt.png (1.51 KiB) Viewed 388 times

, calculate

nu.png (911 Bytes) Viewed 388 times

and select the correct option:

kerute.png (4.2 KiB) Viewed 388 times

how do I solve this?
Your help is deeply appreciated.
Thank you.

[Baltuilhe]

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

[nathi123]

[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]

[Guest]

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?

[Guest]

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

[Guest]

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].