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)=(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.