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Question Number 32701 by caravan msup abdo. last updated on 31/Mar/18

l e t g i v e f (x) = \frac{x}{\sqrt{x + 1}}

1) c a l c u l a t e f^{- 1} (x))

2) c a l c u l a t e {(f^{- 1})}^{^{'}} (x) .

Commented by Rio Mike last updated on 31/Mar/18

solution

let f (x) = y

y = \frac{x}{\sqrt{x + 1}}

\frac{y \sqrt{x + 1}}{y} = \frac{x}{y}

x + 1 = {(\frac{x}{y})}^{2}

x = \frac{x^{2}}{y^{2}} - 1

replacing y by x .

f^{- 1} (x) = (\frac{x^{2_{}}}{x^{2}}) - 1

Commented by MJS last updated on 31/Mar/18

you must exchange x and y

each x must be replaced by y

and each y must be replaced by x

Commented by abdo imad last updated on 01/Apr/18

l e t p u t f (x) = y \Leftrightarrow x = f^{- 1} (y) \Rightarrow \frac{x}{\sqrt{x + 1}} = y \Rightarrow x = y \sqrt{x + 1}

w i t h c o n d i t i o n a x > - 1 a n d \frac{x}{y} > 0 \Rightarrow x^{2} = y^{2} (x + 1) = y^{2} x + y^{2} \Rightarrow

x^{2} - y^{2} x - y^{2} = 0 e q u . w t h u n k n o w n x

Δ = y^{4} + 4 y^{2} ⩾ 0 \Rightarrow x_{1} = \frac{y^{2} + \sqrt{y^{4} + 4 y^{2}}}{2}

x_{2} = \frac{y^{2} - \sqrt{y^{4} + 4 y^{2}}}{2} b u t w e m u s t h a v e x + 1 > 0 l e t t r y

w i t h x_{2} b e c a u s e i t s c l e a r t h a t x_{1} + 1 > 0

x_{2} + 1 = \frac{y^{2} - \sqrt{y^{4} + 4 y^{2}}}{2} + 1 = \frac{y^{2} + 2 - \sqrt{y^{4} + 4 y^{2}}}{2}

{(y^{2} + 2)}^{2} - (y^{4} + 4 y^{2}) = 4 c o n d i t i o n r e a l i z e d

\frac{x_{2}}{y} = \frac{y^{2} - ∣ y ∣ \sqrt{y^{2} + 4}}{2 y} = y + ξ \frac{\sqrt{4 + y^{2}}}{2} w i t h ξ^{2} = 1 b u t t h e s i n e

o f \frac{x_{2}}{y} i s n o t c o n s t a n t s o t h e s o < u t i o n o f t h e e q u . i s x_{1}

a n d f^{- 1} (x) = \frac{1}{2} (x^{2} + \sqrt{x^{4} + 4 x^{2}})

2) {(f^{- 1})}^{^{'}} (x) = \frac{1}{2} (2 x + \frac{4 x^{3} + 8 x}{2 \sqrt{x^{4} + 4 x^{2}}})

= x + \frac{2 x^{3} + 4 x}{\sqrt{x^{4} + 4 x^{2}}} .

Answered by MJS last updated on 31/Mar/18

x = \frac{y}{\sqrt{y + 1}}

x \sqrt{y + 1} = y

x^{2} y + x^{2} = y^{2}

y^{2} - x^{2} y - x^{2} = 0

y = \frac{x^{2}}{2} \pm \sqrt{\frac{x^{4}}{4} + x^{2}}

y = \frac{x^{2}}{2} \pm \sqrt{\frac{x^{2}}{4} (x^{2} + 4)}

y = f^{- 1} (x) = \frac{x^{2}}{2} \pm \frac{x}{2} \sqrt{x^{2} + 4}

y^{'} = \frac{2 x}{2} \pm (\frac{1}{2} \times {(x^{2} + 4)}^{\frac{1}{2}} + \frac{x}{2} \times \frac{1}{2} \times {(x^{2} + 4)}^{- \frac{1}{2}} \times 2 x)

y^{'} = x \pm (\frac{1}{2} {(x^{2} + 4)}^{\frac{1}{2}} + \frac{x^{2}}{2} {(x^{2} + 4)}^{- \frac{1}{2}})

y^{'} = x \pm \frac{1}{2} {(x^{2} + 4)}^{- \frac{1}{2}} ({(x^{2} + 4)}^{1} + x^{2})

y^{'} = {(f^{- 1})}^{'} (x) = x \pm \frac{x^{2} + 2}{\sqrt{x^{2} + 4}}

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