calculus-preliminary-Q-f-x-2-x-2-x-f-1-x-solution-y-2-x-2-x-y-2-2x-1-2-x-2-2x-y2-x-1-0-2-x-t-

Question Number 135215 by mnjuly1970 last updated on 11/Mar/21

\dots . . c a l c u l u s p r e l i m i n a r y \dots .

Q : f (x) = 2^{x} - 2^{- x} \Rightarrow f^{- 1} (x) = ? ? ?

s o l u t i o n :

y = 2^{x} - 2^{- x} \dots . .

y = \frac{2^{2 x} - 1}{2^{x}} \Rightarrow 2^{2 x} - y 2^{x} - 1 = 0 (*) \dots

:: 2^{x} = t \Rightarrow t > 0 \dots ✓ \dots .

(*) \to \dots t^{2} - t y - 1 = 0 \dots .

Δ = y^{2} + 4 > 0 \dots ✓ \dots

t = \frac{y + \sqrt{y^{2} + 4}}{2} \dots \dots

:: 2^{x} = \frac{y + \sqrt{y^{2} + 4}}{2} \underset{b o t h s i d e s}{\overset{t a k i n g l o g}{\Rightarrow}} \dots .

x := {l o g}_{2} (\frac{y + \sqrt{y^{2} + 4}}{2})

f^{- 1} (x) = {l o g}_{2} (\frac{x + \sqrt{x^{2} + 4}}{2}) ✓ ✓

\dots \dots \dots \dots \dots \dots \dots \dots .