# Let f and g be two real functions...

### Let f and g be two real functions...

Let f and g be two real functions such that g is the inverse of f is defined as
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, calculate
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and select the correct option:

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how do I solve this?
Thank you.
Learningforcomp

Posts: 9
Joined: Fri Sep 11, 2020 2:35 pm
Reputation: 0

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

Good afternoon!

$$y=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}$$

$$y=\frac{e^{x}-\frac{1}{e^{x}}}{e^{x}+\frac{1}{e^{x}}}$$

$$y=\frac{\frac{e^{2x}-1}{e^{x}}}{\frac{e^{2x}+1}{e^{x}}}$$

$$y=\frac{e^{2x}-1}{e^{2x}+1}$$

$$y\cdot\left(e^{2x}+1\right)=e^{2x}-1$$

$$y\cdot e^{2x}-e^{2x}=-1-y$$

$$e^{2x}\cdot\left(y-1\right)=-\left(1+y\right)$$

$$e^{2x}=\frac{y+1}{y-1}$$

$$2x=\ln\left(\frac{y+1}{y-1}\right)$$

$$x=g(y)=\frac{1}{2}\cdot\ln\left(\frac{y+1}{y-1}\right)$$

Try to reach the answer, now

Good luck

Baltuilhe

Posts: 64
Joined: Fri Dec 14, 2018 3:55 pm
Reputation: 44

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

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

nathi123

Posts: 27
Joined: Sun Sep 17, 2017 1:56 pm
Reputation: 26

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

Baltuilhe wrote:Good afternoon!

$$y=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}$$

$$y=\frac{e^{x}-\frac{1}{e^{x}}}{e^{x}+\frac{1}{e^{x}}}$$

$$y=\frac{\frac{e^{2x}-1}{e^{x}}}{\frac{e^{2x}+1}{e^{x}}}$$

$$y=\frac{e^{2x}-1}{e^{2x}+1}$$

$$y\cdot\left(e^{2x}+1\right)=e^{2x}-1$$

$$y\cdot e^{2x}-e^{2x}=-1-y$$

$$e^{2x}\cdot\left(y-1\right)=-\left(1+y\right)$$

$$e^{2x}=\frac{y+1}{y-1}$$

$$2x=\ln\left(\frac{y+1}{y-1}\right)$$

$$x=g(y)=\frac{1}{2}\cdot\ln\left(\frac{y+1}{y-1}\right)$$

Try to reach the answer, now

Good luck

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

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

That must have been a typo. $$\frac{-(y+1)}{y-1}= \frac{y+1}{-(y-1)}= \frac{y+1}{1- y}$$.
Guest

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

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

So $$2(e^x- e^{-x})= e^x+ e^{-x}$$
$$2e^x- 2e^{-x}= e^x+ e^{-x}$$

$$e^{x}= 3e^{-x}$$
$$e^{2x}= 3$$
$$2x= ln(3)$$
$$x= \frac{ln(x)}{2}$$.
Guest