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Question Number 39702 by math khazana by abdo last updated on 10/Jul/18

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

1) f i n d D_{f}

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

3) d e t e r m i n e f^{- 1} (x)

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

5) l e t u (x) = x^{2} + 4

d e t e r m i n e v (x) = f o u (x) a n d c a l c u l a t e

v^{^{'}} (x)

6) c a l c u l a t e \int_{3}^{5} f (x) d x .

Commented by math khazana by abdo last updated on 13/Jul/18

1) x \in D_{f} \Leftrightarrow x ⩾ 3 a n d x + 2 - \sqrt{x - 3} ⩾ 0 \Rightarrow

x ⩾ 3 a n d {(x + 2)}^{2} ⩾ x - 3 \Rightarrow x^{2} + 4 x + 4 - x + 3 ⩾ 0 \Rightarrow

x^{2} + 3 x + 7 ⩾ 0 t h i s i n e q u a l i t y i s t r u e b e c a u s e

Δ = 9 - 28 < 0 \Rightarrow D_{f} = [3, + \infty [

2) w e h a v e f^{2} (x) = 4 (x - \sqrt{x - 3} + 2) \Rightarrow

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

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

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

Commented by math khazana by abdo last updated on 13/Jul/18

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

\Rightarrow x^{2} = 4 (y + 2 - \sqrt{y - 3}) \Rightarrow

x^{2} = 4 y + 8 - 4 \sqrt{y - 3} \Rightarrow 4 \sqrt{y - 3} = 4 y + 8 - x^{2} \Rightarrow

16 (y - 3) = {(4 y + 8 - x^{2})}^{2} \Rightarrow

16 y - 48 = {(4 y + 8)}^{2} - 2 x^{2} (4 y + 8) + x^{4}

16 y - 48 = 16 y^{2} + 64 y + 64 - 8 {y x}^{2} - 16 x^{2} + x^{4} \Rightarrow

x^{4} - 8 (y + 2) x^{2} + 16 y^{2} + 64 - 16 y + 48 = 0 \Rightarrow

x^{4} - 8 (y + 2) x^{2} + 16 y^{2} - 16 y + 112 = 0

Δ^{^{'}} = {(- 2 (y + 2))}^{2} - 16 y^{2} + 16 y - 112

= 4 (y^{2} + 4 y + 4) - 16 y^{2} + 16 y - 112

= - 8 y^{2} + 32 y + 16 - 112 \dots b e c o n t i n u e d \dots

Commented by maxmathsup by imad last updated on 14/Jul/18

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

= 2 \sqrt{x^{2} + 4 - \sqrt{x^{2} + 1} + 2}

= 2 \sqrt{x^{2} - \sqrt{x^{2} + 1} + 6}

w e h a v e v^{2} (x) = 4 {x^{2} - \sqrt{x^{2} + 1} + 2} \Rightarrow

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

v (x) v^{^{'}} (x) = 2 {\frac{2 x \sqrt{x^{2} + 1} - x}{\sqrt{x^{2} + 1}}} \Rightarrow v^{^{'}} (x) = \frac{2}{v (x)} \frac{2 x \sqrt{x^{2} + 1} - x}{\sqrt{x^{2} + 1}}

= \frac{2 x \sqrt{x^{2} + 1} - x}{(\sqrt{x^{2} + 6 - \sqrt{x^{2} + 1}) (\sqrt{x^{2} + 1)}}} .

Commented by maxmathsup by imad last updated on 14/Jul/18

6) \int_{3}^{5} f (x) d x = 2 \int_{3}^{5} \sqrt{x + 2 - \sqrt{x - 3}} d x c h a n g e m e n t \sqrt{x - 3} = t g i v e x = 3 + t^{2}

\int_{3}^{5} f (x) d x = 2 \int_{0}^{\sqrt{2}} \sqrt{5 + t^{2} - t} 2 t d t = 4 \int_{0}^{\sqrt{2}} t \sqrt{t^{2} - t + 5} d t

= 4 \int_{0}^{\sqrt{2}} t \sqrt{{(t - \frac{1}{2})}^{2} + 5 - \frac{1}{4}} d t = 4 \int_{0}^{\sqrt{2}} t \sqrt{{(t - \frac{1}{2})}^{2} + \frac{19}{4}} d t

=_{t - \frac{1}{2} = \frac{\sqrt{19}}{2} s h (u)} 4 \int_{- a r s h (\frac{1}{\sqrt{19}})}^{a r g s h (\frac{2 \sqrt{2} - 1}{\sqrt{19}})} (\frac{1}{2} + \frac{\sqrt{19}}{2} s h (u)) c h (u) \frac{\sqrt{19}}{2} c h (u) d u

= \sqrt{19} \int_{- a r g s h (\frac{1}{\sqrt{19}})}^{a r g s h (\frac{2 \sqrt{2} - 1}{\sqrt{19}})} (1 + \sqrt{19} s h (u)) c h (u) d u

= \sqrt{19} \int_{- a r g s h (\frac{1}{\sqrt{19}})}^{a r g s h (\frac{2 \sqrt{2} - 1}{\sqrt{19}})} c h (u) d u + \sqrt{19} \int_{- a r g s h (\frac{1}{\sqrt{19}})}^{a r g s h (\frac{2 \sqrt{2} - 1}{\sqrt{19}})} s h (u) c h (u) d u

= \sqrt{19 {} \frac{2 \sqrt{2} - 1}{\sqrt{19}} + \frac{1}{\sqrt{19}}} + \frac{\sqrt{19}}{2} {{(\frac{2 \sqrt{2} - 1}{\sqrt{19}})}^{2} - (\frac{1}{19})} .

Commented by ajfour last updated on 14/Jul/18

G r e a t w o r k s i r!

Commented by math khazana by abdo last updated on 15/Jul/18

t h a n k y o u s i r .