f-1-x-f-x-f-x-

Question Number 2075 by 123456 last updated on 01/Nov/15

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

f (x) = ? ?

Commented by Yozzi last updated on 01/Nov/15

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

\Rightarrow \int 1 d x = \int f (x) f^{^{'}} (x) d x

x + c = \frac{1}{2} {[f (x)]}^{2}

2 x + c = {[f (x)]}^{2}

f (x) = \pm \sqrt{2 x + c}

e . g, c = 0, f (x) > 0 \Rightarrow f (x) = \sqrt{2 x}

∴ f^{^{'}} (x) = \sqrt{2} \times \frac{1}{2 \sqrt{x}}

f^{^{'}} (x) = \frac{1}{\sqrt{2 x}} = \frac{1}{f (x)}

Commented by prakash jain last updated on 01/Nov/15

f^{- 1} (x) is for inverse of f ?

Commented by 123456 last updated on 01/Nov/15

yes, inverse function

Commented by Rasheed Soomro last updated on 02/Nov/15

O n t h e c o m m e n t o f Y o z z i

W h y f (x) > 0 [i n 6^{t h} l i n e]

I s f (x) \neq 0 n o t s u f f i c i e n t ?

∴ f (x) = \pm \sqrt{2 x} [N o t o n l y + \sqrt{2 x}]

Commented by Yozzi last updated on 02/Nov/15

I w a s g i v i n g a n e x a m p l e .

e x a m p l e \equiv e . g .

T h e i n t e r p r e t a t i o n o f t h e q u e s t i o n

w a s i n c o r r e c t .

Answered by prakash jain last updated on 01/Nov/15

f (x) = {k x}^{n}

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

y = {k x}^{n} \Rightarrow x = {(\frac{y}{k})}^{1 / n} \Rightarrow f^{- 1} (x) = {(\frac{x}{k})}^{1 / n}

\frac{1}{n} = n - 1 \Rightarrow n^{2} - n - 1 = 0

n = \frac{1 \pm \sqrt{5}}{2} \dots . . (1)

n k = \frac{1}{k^{1 / n}} \Rightarrow n k = k^{- 1 / n} \Rightarrow k^{- \frac{1 + n}{n}} = n \Rightarrow k = n^{- \frac{n}{n + 1}} \dots . (2)

f (x) = {k x}^{n} should satisfy the given equation

where n and k are given in (1) and (2) .

Checks to be done .