If-f-x-2x-2-5-f-1-2-

Question Number 150187 by mathdanisur last updated on 10/Aug/21

If f (x) = {2 x}^{2} + 5 \Rightarrow f^{- 1} (2) = ?

Commented by amin96 last updated on 10/Aug/21

2 x^{2} + 5 = 2 \Rightarrow x = \pm \frac{3 i}{\sqrt{2}}

f (\pm \frac{3 i}{\sqrt{2}}) = 2 \Rightarrow f^{- 1} (2) = \pm \frac{3 i}{\sqrt{2}}

Answered by puissant last updated on 10/Aug/21

f (x) = y

\Rightarrow {2 x}^{2} + 5 = y

\Rightarrow {2 x}^{2} = y - 5

\Rightarrow x = \pm \sqrt{\frac{1}{2} (y - 5)}

f^{- 1} (x) = \pm \frac{\sqrt{2}}{2} \sqrt{x - 5} \dots

Commented by puissant last updated on 10/Aug/21

thank you sir

Commented by Rasheed.Sindhi last updated on 10/Aug/21

P e r h a p s {t h e r e}^{'} s a t y p o s i r

x = \pm \sqrt{\frac{1}{2} (y - 5)} \Rightarrow f^{- 1} (x) = \pm \frac{\sqrt{2 (x - 5)}}{2}

f^{- 1} (2) will not be real .

Commented by Rasheed.Sindhi last updated on 10/Aug/21

{you}^{'} r e welcome sir

Commented by mathdanisur last updated on 10/Aug/21

thankyou Ser

Answered by liberty last updated on 10/Aug/21

let f^{- 1} (2) = t \Rightarrow f (t) = 2

\Rightarrow {2 t}^{2} + 5 = 2

\Rightarrow t^{2} = - \frac{3}{2}; t = \pm i \sqrt{\frac{3}{2}}

Commented by mathdanisur last updated on 10/Aug/21

t h a n k y o u S e r