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Question Number 46419 by maxmathsup by imad last updated on 25/Oct/18

l e t f (x) = \sqrt{x + 1 - \sqrt{x - 1}}

1) f i n d D_{f}

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

3) f i n d \int f (x) d x

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

5) l e t g (x) = {(c h (x))}^{2} c a l c u l a t e f o g (x) a n d {(f o g)}^{^{'}} (x) .

Commented by maxmathsup by imad last updated on 28/Oct/18

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

\Rightarrow {(x + 1)}^{2} ⩾ x - 1 a n d x ⩾ 1 \Rightarrow x^{2} + 2 x + 1 - x + 1 ⩾ 0 a n d x ⩾ 1

\Rightarrow x (x + 1) ⩾ 0 a n d x ⩾ 1 \Rightarrow D_{f} = [1, + \infty [

2) l e t f (x) = y \Rightarrow x = f^{- 1} (y)

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

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

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

x^{2} + (1 - 2 y^{2}) x + y^{4} - 2 y^{2} + 2 = 0 t h i s e q u a t i o n h a v e s^{t} \Leftrightarrow Δ ⩾ 0 \Rightarrow

{(1 - 2 y^{2})}^{2} - 4 (y^{4} - 2 y^{2} + 2) ⩾ 0 \Rightarrow 4 y^{4} - 4 y^{2} + 1 - 4 y^{4} + 8 y^{2} - 8 ⩾ 0 \Rightarrow

4 y^{2} - 7 ⩾ 0 s o x_{1} = \frac{2 y^{2} - 1 + \sqrt{4 y^{2} - 7}}{2} a n d x_{2} = \frac{2 y^{2} - 1 - \sqrt{4 y^{2} - 7}}{2}

f^{- 1} (x) = \frac{2 x^{2} - 1 \bar{+} \sqrt{4 x^{2} - 7}}{2}

Commented by maxmathsup by imad last updated on 28/Oct/18

3) \int f (x) d x = \int \sqrt{x - 1 + 2 - \sqrt{x - 1}} d x c h a n g e m e n t \sqrt{x - 1} = t g i v e x - 1 = t^{2} \Rightarrow

\int f (x) d x = \int \sqrt{t^{2} + 2 - t} 2 t d t = 2 \int t \sqrt{t^{2} - t + 2} d t

= 2 \int t \sqrt{t^{2} - 2 \frac{1}{2} t + \frac{1}{4} + 2 - \frac{1}{4}} d t = 2 \int t \sqrt{{(t - \frac{1}{2})}^{2} + \frac{7}{4}} d t

=_{t - \frac{1}{2} = \frac{\sqrt{7}}{2} s h (t)} 2 \int (\frac{\sqrt{7}}{2} s h (t) + \frac{1}{2}) \frac{\sqrt{7}}{2} c h (t) \frac{\sqrt{7}}{2} c h (t) d t

= \frac{7}{4} \int (1 + \sqrt{7} s h (t) {c h}^{2} (t) d t

= \frac{7}{4} \int {c h}^{2} t d t + \frac{7 \sqrt{7}}{4} \int s h (t) {c h}^{2} (t) d t b u t \int {c h}^{2} t d t = \int \frac{1 + c h (2 t)}{2} d t

= \frac{t}{2} + \frac{1}{4} s h (2 t) = \frac{t}{2} + \frac{1}{2} s h (t) c h (t) a n d \int s h (t) {c h}^{2} t d t = \frac{1}{3} {c h}^{3} t \Rightarrow

\int f (x) d x = \frac{7}{8} t + \frac{7}{8} s h t) c h (t) + \frac{7 \sqrt{7}}{12} {c h}^{3} (t) + c

= \frac{7}{8} \sqrt{x - 1} + \frac{7}{8} s h (\sqrt{x - 1)} c h (\sqrt{x - 1}) + \frac{7 \sqrt{7}}{12} {c h}^{3} (\sqrt{x - 1}) + c .

Commented by maxmathsup by imad last updated on 28/Oct/18

4) w e h a v e 2 \int f^{- 1} (x) d x = \int (2 x^{2} - 1) d x \bar{+} \int \sqrt{4 x^{2} - 7} d x b u t

\int (2 x^{2} - 1) d x = \frac{2}{3} x^{3} - x + c_{1}

\int \sqrt{4 x^{2} - 7} d x = \int 2 \sqrt{x^{2} - \frac{7}{4}} d x =_{x = \frac{\sqrt{7}}{2} c h (t)} 2 \int \frac{\sqrt{7}}{2} s h (t) \frac{\sqrt{7}}{2} s h (t) d t

= \frac{7}{2} \int {s h}^{2} (t) d t = \frac{7}{4} \int (c h (2 t) - 1) d t

= \frac{7}{8} s h (2 t) - \frac{7 t}{4} + c_{2}

= \frac{7}{4} s h (t) c h (t) - \frac{7}{4} t + c_{2} = \frac{7}{4} \sqrt{{(\frac{2 x}{\sqrt{7}})}^{2} - 1} \frac{2 x}{\sqrt{7}} - \frac{7}{4} a r g c h (\frac{2 x}{\sqrt{7}}) + c_{2}

= \frac{\sqrt{7}}{2} \sqrt{\frac{4 x^{2}}{7} - 1} - \frac{7}{4} l n (\frac{2 x}{\sqrt{7}} + \sqrt{\frac{4 x^{2}}{7} - 1}) + c_{2} \Rightarrow

\int f^{- 1} (x) d x = \frac{x^{3}}{3} - \frac{x}{2} \bar{+} {\frac{\sqrt{7}}{4} \sqrt{\frac{4 x^{2}}{7} - 1} - \frac{7}{8} l n (\frac{2 x}{\sqrt{7}} + \sqrt{\frac{4 x^{2}}{7} - 1}) + C .

Commented by maxmathsup by imad last updated on 28/Oct/18

e r r o r o f t y p o

\int f^{- 1} (x) d x = \frac{x^{3}}{3} - \frac{x}{2} \bar{+} {\frac{\sqrt{7}}{4} x \sqrt{\frac{4 x^{2}}{7} - 1} - \frac{7}{8} l n (\frac{2 x}{\sqrt{7}} + \sqrt{\frac{4 x^{2}}{7} - 1}) + C