Question

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​

Answer 1

We're asked to find inverse of the function,

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

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

$\longrightarrow\rm{y=\dfrac{e^x-e^{-x}}{e^x+e^{-x}}+2}$

Subtract 2.

$\longrightarrow\rm{y-2=\dfrac{e^x-e^{-x}}{e^x+e^{-x}}}$

We apply componendo - dividendo rule.

$\rm{\dfrac{a}{b}=\dfrac{c}{d}\quad\iff\quad\dfrac{b+a}{b-a}=\dfrac{d+c}{d-c}}$

Thus,

$\longrightarrow\rm{\dfrac{y-2}{1}=\dfrac{e^x-e^{-x}}{e^x+e^{-x}}}$

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

$\longrightarrow\rm{\dfrac{y-1}{3-y}=\dfrac{2e^x}{2e^{-x}}}$

$\longrightarrow\rm{\dfrac{e^x}{e^{-x}}=\dfrac{y-1}{3-y}}$

$\longrightarrow\rm{e^{2x}=\dfrac{y-1}{3-y}}$

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

Take square root.

$\longrightarrow\rm{e^x=\sqrt{\dfrac{y-1}{3-y}}}$

Take natural logarithm.

$\longrightarrow\rm{x=\log_e\sqrt{\dfrac{y-1}{3-y}}}$

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

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

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

Replace y by x and finally we get,

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

This is the inverse of the given function.