The inverse of the function f (x) = e^x-e^- x/e^x+e^- x + 2 is given by a.log (x - 2/x - 1 )^1 2 b.12loge (x - 1/3 - x ) c.&log (x/2 - x )^1 2 & d.log (x - 1/3 - x )^1 2…

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asked Mar 5, 2023 233k views

1 Answer

Sagar Damani · Answer 1 · 2023-03-09T17:22:46+0000

We're asked to find inverse of the function,

$\longrightarrow\rm{f(x)=(e^x-e^(-x))/(e^x+e^(-x))+2}$

Take
$\rm{y=f(x).}$

$\longrightarrow\rm{y=(e^x-e^(-x))/(e^x+e^(-x))+2}$

Subtract 2.

$\longrightarrow\rm{y-2=(e^x-e^(-x))/(e^x+e^(-x))}$

We apply componendo - dividendo rule.

$\rm{(a)/(b)=(c)/(d)\quad\iff\quad(b+a)/(b-a)=(d+c)/(d-c)}$

Thus,

$\longrightarrow\rm{(y-2)/(1)=(e^x-e^(-x))/(e^x+e^(-x))}$

$\Longrightarrow\rm{(1+(y-2))/(1-(y-2))=((e^x+e^(-x))+(e^x-e^(-x)))/((e^x+e^(-x))-(e^x-e^(-x)))}$

$\longrightarrow\rm{(y-1)/(3-y)=(2e^x)/(2e^(-x))}$

$\longrightarrow\rm{(e^x)/(e^(-x))=(y-1)/(3-y)}$

$\longrightarrow\rm{e^(2x)=(y-1)/(3-y)}$

$\longrightarrow\rm{(e^x)^2=(y-1)/(3-y)}$

Take square root.

$\longrightarrow\rm{e^x=\sqrt{(y-1)/(3-y)}}$

Take natural logarithm.

$\longrightarrow\rm{x=\log_e\sqrt{(y-1)/(3-y)}}$

$\longrightarrow\rm{x=(1)/(2)\log_e\left((y-1)/(3-y)\right)}$

Now take
$\rm{x=f^(-1)(y).}$

$\longrightarrow\rm{f^(-1)(y)=(1)/(2)\log_e\left((y-1)/(3-y)\right)}$

Replace y by x and finally we get,

$\longrightarrow\rm{\underline{\underline{f^(-1)(x)=(1)/(2)\log_e\left((x-1)/(3-x)\right)}}}$

This is the inverse of the given function.