The inverse function of
f(x) = ${{{8^{2x}} - {8^{ - 2x}}} \over {{8^{2x}} + {8^{ - 2x}}}}$, x $\in$ (-1, 1), is :
Solution
f(x) = ${{{8^{2x}} - {8^{ - 2x}}} \over {{8^{2x}} + {8^{ - 2x}}}}$ = y
<br><br>$\therefore$ ${{y + 1} \over {y - 1}} = {{{{2.8}^{2x}}} \over { - {{2.8}^{ - 2x}}}}$
<br><br>$\Rightarrow$ ${{1 + y} \over {1 - y}}$ = 8<sup>4x</sup>
<br><br>$\Rightarrow$ ${\log _e}\left( {{{1 + y} \over {1 - y}}} \right)$ = 4x ${\log _e}8$
<br><br>$\Rightarrow$ x = ${1 \over {4{{\log }_e}8}}{\log _e}\left( {{{1 + y} \over {1 - y}}} \right)$
<br><br>$\therefore$ f<sup>-1</sup>(x) = $${1 \over 4}\left( {{{\log }_8}e} \right){\log _e}\left( {{{1 + x} \over {1 - x}}} \right)$$
About this question
Subject: Mathematics · Chapter: Sets, Relations and Functions · Topic: Sets and Operations
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