solve-for-f-x-such-that-f-x-f-1-x-

Question Number 181541 by mr W last updated on 26/Nov/22

s o l v e f o r f (x) s u c h t h a t

f^{'} (x) = f^{- 1} (x)

Commented by a.lgnaoui last updated on 26/Nov/22

{[f^{- 1} (x)]}^{^{'}} = \frac{1}{f^{^{'}} (x) f (x)} f^{'} (x) = f^{- 1} (x)

{[f^{- 1} (x)]}^{^{'}} = \frac{1}{f^{- 1} (x) f (x)}

f^{- 1} (x) {[f^{- 1} (x)]}^{^{'}} = \frac{1}{f (x)}

2 \times {({[f^{- 1} (x)]}^{2})}^{^{'}} = \frac{1}{f (x)} (2)

\Rightarrow ({[{(f^{^{'}} (x)]}^{2})}^{^{'}} = \frac{1}{2 f (x)}

Prime causes double exponent: use braces to clarify

\dots \dots .

Commented by universe last updated on 26/Nov/22

s o l v e f o r f (x) s u c h t h a t

f^{'} (x) = f^{- 1} (x)

Answered by a.lgnaoui last updated on 26/Nov/22

f^{'} (x) \times f (x) = 1

{[\frac{{(f (x))}^{2}}{2}]}^{^{'}} = 1 \Rightarrow \frac{{[f (x)]}^{2}}{2} = x + C

f (x) = \sqrt{2 x + k} k \in R

Commented by a.lgnaoui last updated on 26/Nov/22

o h y e s

Commented by cortano1 last updated on 26/Nov/22

f^{- 1} (x) \neq \frac{1}{f (x)}

Answered by cortano1 last updated on 26/Nov/22

\frac{df (x)}{dx} = f^{- 1} (x)

f (x) = \int f^{- 1} (x) dx

[let x = f (u) \Rightarrow dx = df (u)]

f (x) = \int f^{- 1} (f (u)) df (u)

f (x) = \int u d (f (u))

f (x) = u f (u) - \int f (u) du

f (x) = x f^{- 1} (x) - F (f (x)) + c

Commented by mr W last updated on 26/Nov/22

w e s h o u l d d e t e r m i n e f (x) .

Answered by universe last updated on 26/Nov/22

s e e q u e s t i o n 173624

Answered by mahdipoor last updated on 26/Nov/22

\frac{d f}{d x} \times \frac{{d f}^{- 1}}{d x} = 1 \Rightarrow \frac{d f}{d x} = f^{- 1} = \frac{d x}{{d f}^{- 1}} \Rightarrow

\Rightarrow f^{- 1} {d f}^{- 1} = d x \Rightarrow \frac{1}{2} {(f^{- 1} (x))}^{2} = x + c

f^{- 1} (x) = {[2 (x + c)]}^{1 / 2} \Rightarrow x = f ({[2 (x + c)]}^{1 / 2})