let-give-f-x-x-2-x-1-1-find-f-1-x-inverse-of-f-x-2-calculate-f-1-x-

Question Number 32283 by abdo imad last updated on 22/Mar/18

l e t g i v e f (x) = x + 2 - \sqrt{x + 1}

1) f i n d f^{- 1} (x) i n v e r s e o f f (x)

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

Commented by abdo imad last updated on 24/Mar/18

D_{f} = [- 1, + \infty [f (x) = y \Leftrightarrow y = x + 2 - \sqrt{x + 1} \Leftrightarrow

{(x + 2 - y)}^{2} = x + 1 \Rightarrow {(x + 2)}^{2} - 2 y (x + 2) + y^{2} - x - 1 = 0

\Rightarrow x^{2} + 4 x + 4 - 2 y x - 4 y + y^{2} - x - 1 = 0 \Rightarrow

x^{2} + (3 - 2 y) x + y^{2} - 4 y + 3 = 0

Δ = {(3 - 2 y)}^{2} - 4 (y^{2} - 4 y + 3) =

= 9 - 12 y + 4 y^{2} - 4 y^{2} + 16 y - 12 = 4 y - 3

x_{1} = \frac{- 3 + 2 y + \sqrt{4 y - 3}}{2} a n d x_{2} = \frac{- 3 + 2 y - \sqrt{4 h - 3}}{2}

w e m u s t h a v e x + 2 - y ⩾ 0 l e t v e r i f y

x_{2} + 2 - y = \frac{- 3 + 2 y - \sqrt{4 y - 3}}{2} + 2 - y

= \frac{- 3 + 2 y - \sqrt{4 y - 3} + 4 - 2 y}{2} = \frac{1 - y - \sqrt{4 y - 3}}{2} < 0 s o t h e

s o u t i o n i s x_{1} a n d f^{- 1} (x) = \frac{2 x + \sqrt{4 x - 3} - 3}{2}

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

{(f^{- 1})}^{^{'}} (x) = 1 + \frac{1}{2} \frac{4}{2 \sqrt{4 x - 3}} = 1 + \frac{1}{\sqrt{4 x - 3}} . w i t h x > \frac{3}{4} .

Answered by MJS last updated on 22/Mar/18

x = y + 2 - \sqrt{y + 1}

\sqrt{y + 1} = - x + y + 2

y + 1 = y^{2} + 2 (2 - x) y + {(2 - x)}^{2}

y^{2} + (3 - 2 x) y + (x^{2} - 4 x + 3) = 0

p = 3 - 2 x; q = (x^{2} - 4 x + 3)

y = - \frac{p}{2} \pm \frac{\sqrt{p^{2} - 4 q}}{2}

y = x - \frac{3}{2} \pm \frac{\sqrt{4 x - 3}}{2}

y^{'} = 1 \pm \frac{\frac{1}{2} {(4 x - 3)}^{- \frac{1}{2}} \times 4}{2}

y^{'} = 1 \pm \frac{1}{\sqrt{4 x - 3}}

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