Question

find the invers of this function:
$f(x)=(2x+3)/\sqrt{x^2-1}$

Answer 1

Step-by-step explanation:

We have,

$\rm \: f(x) = \frac{2x + 3}{ \sqrt{ {x}^{2} - 1} }$

Let $\rm\:y=f^{-1}(x)\implies\:f(y)=x$

So ,

$\rm \: f(y) = \frac{2y+ 3}{ \sqrt{ {y}^{2} - 1} } = x$

$\rm \: \implies \frac{2y+ 3}{ \sqrt{ {y}^{2} - 1} } = x$

$\rm \: \implies \bigg( \frac{2y+ 3}{ \sqrt{ {y}^{2} - 1} } \bigg)^{2} = x^{2}$

$\rm \: \implies \frac{(2y+ 3)^{2} }{ {y}^{2} - 1} = x^{2}$

$\rm \: \implies \frac{4y^{2} + 12y + 9}{ {y}^{2} - 1} = x^{2}$

$\rm \: \implies 4y^{2} + 12y + 9= x^{2} ( {y}^{2} - 1)$

$\rm \: \implies 4y^{2} + 12y + 9= x^{2} {y}^{2} - {x}^{2}$

$\rm \: \implies ( 4 - {x}^{2}) y^{2} + 12y + (9 + {x}^{2}) = 0$

$\rm \: \implies \: y = \frac{ - 12 \pm \sqrt{(12) ^{2} - 4(4 - {x}^{2} )(9 + {x}^{2} )}}{2(4 - {x}^{2}) }$

$\rm \: \implies \: y = \frac{ - 12 \pm \sqrt{144 + 4( {x}^{2} - 4 )( {x}^{2} + 9) } }{2(4 - {x}^{2}) }$

$\rm \: \implies \: y = \frac{ - 12 \pm \sqrt{144 + 4 \{ {x}^{4} + 9 {x}^{2} - 4 {x}^{2} - 36 \} } }{2(4 - {x}^{2}) }$

$\rm \: \implies \: y = \frac{ - 12 \pm \sqrt{144 + 4 ( {x}^{4} + 5 {x}^{2} - 36 ) } }{2(4 - {x}^{2}) }$

$\rm \: \implies \: y = \frac{ - 12 \pm \sqrt{144 + 4 {x}^{4} + 20 {x}^{2} - 144} }{2(4 - {x}^{2}) }$

$\rm \: \implies \: y = \frac{ - 12 \pm \sqrt{ 4 {x}^{4} + 20 {x}^{2} } }{2(4 - {x}^{2}) }$

$\rm \: \implies \: y = \frac{ - 12 \pm \sqrt{ 4 {x}^{2}( {x}^{2} + 5) } }{2(4 - {x}^{2}) }$

$\rm \: \implies \: y = \frac{ - 12 \pm2x \sqrt{ {x}^{2} + 5 } }{2(4 - {x}^{2}) }$

$\rm \: \implies \: y = \frac{ - 6 \pm x \sqrt{ {x}^{2} + 5 } }{4 - {x}^{2}}$

$\rm \: \implies \: y = \frac{ 6 \mp x \sqrt{ {x}^{2} + 5 } }{ {x}^{2} - 4}$

so,

We have,

$\rm \: \implies \: f ^{ - 1}(x) = \frac{ 6 + | x \sqrt{ {x}^{2} + 5 } | }{ {x}^{2} - 4}$