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Question Number 34736 by math khazana by abdo last updated on 10/May/18

l e t f (x) = - 2 x + \sqrt{x - 3}

1) f i n d f^{- 1} (x)

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

3) l e t g (x) = x^{2} - 2 x + 3

c a l c u l a t e f o g (x) a n d g i v e D_{f o g}

4) f i n d {(f o g)}^{- 1} (x)

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

Commented by prof Abdo imad last updated on 13/May/18

D_{f} = [- 3, + \infty [l e t p u t y = f (x) \Leftrightarrow x = f^{- 1} (y)

y = f (x) \Leftrightarrow y = - 2 x + \sqrt{x - 3} \Leftrightarrow {(y + 2 x)}^{2} = x - 3

\Rightarrow y^{2} + 4 y x + 4 x^{2} - x + 3 = 0 \Rightarrow

4 x^{2} + (4 y - 1) x + y^{2} + 3 = 0

Δ = {(4 y - 1)}^{2} - 16 (y^{2} + 3)

= 16 y^{2} - 8 y + 1 - 16 y^{2} - 48 = - 8 y - 47

= - 8 (y + \frac{47}{8}) Δ m u s t b e ⩾ 0 \Rightarrow y ⩽ - \frac{47}{8}

a n d y + 2 x ⩾ 0 x_{1} = \frac{1 - 4 y + \sqrt{- 8 y - 47}}{8}

x_{2} = \frac{1 - 4 y - \sqrt{- 8 y - 47}}{8}

y + 2 x_{1} = y + \frac{1 - 4 y + \sqrt{- 8 y - 47}}{4}

= \frac{1 + \sqrt{- 8 y - 47}}{4} ⩾ 0 \Rightarrow f^{- 1} (x) = \frac{1 - 4 x + \sqrt{- 8 x - 47}}{8}

w i t h {D f}^{- 1} =] - \infty, - \frac{47}{8}]

2) {(f^{- 1})}^{^{'}} (x) = - \frac{1}{2} + \frac{1}{8} \frac{- 8}{2 \sqrt{- 8 x - 47}}

= - \frac{1}{2} - \frac{1}{\sqrt{- 8 x - 47}} b u t 2 \notin] + \infty, - \frac{47}{8} [s o

{(f^{- 1})}^{^{'}} (2) d o n t e x i s t

3) w e h a v e f (x) = - 2 x + \sqrt{x - 3} a n d

g (x) = x^{2} - 2 x + 3 \Rightarrow

f o g (x) = - 2 g (x) + \sqrt{g (x) - 3}

= - 2 x^{2} + 4 x - 6 + \sqrt{x^{2} - 2 x + 3 - 3}

= - 2 x^{2} + 4 x - 6 + \sqrt{x^{2} - 2 x}

D_{f o g} =] - \infty, 0] \cup [2, + \infty [