Question

Find the inverse of given functions :-

$(a) \: f(x) = log_{a}(x + \sqrt{ {x }^{2} + 1 } )$

$(b) \: y = \frac{ {10}^{x} - {10}^{ - x} }{ {10}^{x} + {10}^{ - x} } + 1$

✔️✔️ quality answer needed✔️✔️

❌❌NO SPAMMING❌❌​

Answer 1

Hope this solution will help you mark as brainliest

Answer 2

ANSWER

$a) f^{-1}(x) = \dfrac{a^{x} + a^{-x} }{2}$

$b)f^{-1}(x) = log_{10}\sqrt{\dfrac{x}{2 - x} }$

Step By Step Explanation

a)Here, we have >

$(a) \: f(x) = log_{a}(x + \sqrt{ {x }^{2} + 1 } )$

Now,

$log_{a}(x + \sqrt{x^{2} + 1}) = y$

So, $x + \sqrt{x^{2} + 1} =a^{y} --------(i)$

$x - \sqrt{x^{2} + 1}=a^{-y} --------(i)(Conjugate)$

On adding (i) and (ii) we get ->

$2x = a^{y} + a^{-y}$

$\implies x = \dfrac {a^{y} + a^{-y} }{2}$

But we know that >>

f(x ) = y and

$x = f^{-1}(y)$

Now,

$f^{-1}(y) =\dfrac{a^{y} + a^{-y} }{2}$

$\implies f^{-1}(x) =\dfrac{a^{x} + a^{-x} }{2}$

b)Here we have >>

$\: y = \dfrac{ {10}^{x} - {10}^{ - x} }{ {10}^{x} + {10}^{ - x} } + 1$

$\implies \: y = \dfrac{ 2 \: .\: {10}^{x} }{ {10}^{x} + {10}^{ - x} } + 1$

$\implies \: y = \dfrac{ 2\: .\: {10}^{x} }{ {10}^{x} + {10}^{ - x} }$

$\implies \: y = \dfrac{ 2\: .\: {10}^{x} }{ {10}^{2x} + 1 }$

$\implies \:10^{2x}y + y = 2\: . \:{10}^{x}$

$\implies 10^{2x}(y-2)= -y$

$\implies 10^{2x}=\dfrac{-y}{y-2}$

$\implies 10^{2x}=\dfrac{y}{2-y}$

$\implies log_{10}(10^{2x}) =log_{10}\dfrac{y}{2-y}$

$\implies {2x} =log_{10}\dfrac{y}{2-y}$

$\implies x=log_{10}\sqrt{\dfrac{y}{2-y} }$

$\implies f^{-1}(y) =log_{10} \sqrt{\dfrac{y}{2-y} }$

$\implies f^{-1}(x) =log_{10} \sqrt{\dfrac{x}{2-x} }$