let f(x)=2(√(x−1)) −2x 1) find D_f 2) study the variation of f(x) 3 ) calculate ∫_1 ^3 f(x)dx 4) determine f^(−1) (x) and calculate ∫_1 ^3 f^(−1) (x)dx 5) find the values of A = ∫_1 ^3 ((f(x))/(f^(−1) (x)dx)) and B = ((∫_1 ^3 f(x))/(∫


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Question Number 42631 by maxmathsup by imad last updated on 29/Aug/18

$l e t f (x) = 2 \sqrt{x - 1} - 2 x$ $1) f i n d D_{f}$ $2) s t u d y t h e v a r i a t i o n o f f (x)$ $3) c a l c u l a t e \int_{1}^{3} f (x) d x$ $4) d e t e r m i n e f^{- 1} (x) a n d c a l c u l a t e \int_{1}^{3} f^{- 1} (x) d x$ $5) f i n d t h e v a l u e s o f A = \int_{1}^{3} \frac{f (x)}{f^{- 1} (x) d x} a n d$ $B = \frac{\int_{1}^{3} f (x)}{\int_{1}^{3} f^{- 1} (x)} d x .$
Commented by bbnnftm833 last updated on 30/Aug/18

$w e n e e d t h e s o l u t i o n$
Commented by maxmathsup by imad last updated on 31/Aug/18

$1) x \in D_{f} \Leftrightarrow x - 1 ⩾ 0 \Rightarrow D_{f} = [1, + \infty [$ $2) {l i m}_{x \to + \infty} f (x) = {l i m}_{x \to + \infty} 2 x (\sqrt{\frac{x - 1}{x^{2}}} - 1) = {l i m}_{x \to + \infty} 2 x (\sqrt{\frac{1}{x} - \frac{1}{x^{2}}} - 1)$ $= {l i m}_{x \to + \infty} - 2 x = - \infty$ $w e h a v e f^{^{'}} (x) = \frac{1}{\sqrt{x - 1}} - 2 f o r x > 1 \Rightarrow f^{^{'}} (x) = \frac{1 - 2 \sqrt{x - 1}}{\sqrt{x - 1}}$ $f^{^{'}} (x) = 0 \Leftrightarrow 2 \sqrt{x - 1} = 1 \Leftrightarrow \sqrt{x - 1} = \frac{1}{2} \Leftrightarrow x - 1 = \frac{1}{4} \Leftrightarrow x = \frac{5}{4}$ $f^{^{'}} (x) > 0 \Leftrightarrow 1 - 2 \sqrt{x - 1} > 0 \Leftrightarrow 2 \sqrt{x - 1} < 1 \Leftrightarrow 1 < x < \frac{5}{4}$ $f^{^{'}} (x) < 0 \Leftrightarrow x > \frac{5}{4}$ $x_{} 1 \frac{5}{4} + \infty$ $f^{^{'}} (x) + -$ $f (x) - 2 i n c - \frac{3}{2} d e c r f (1) = - 2 f (\frac{5}{4}) = 2 \sqrt{\frac{5}{4} - 1} - 2 \frac{5}{4}$ $= 1 - \frac{5}{2} = - \frac{3}{2}$
Commented by maxmathsup by imad last updated on 31/Aug/18
$3) ∫_1 ^3 f(x)dx = ∫_1 ^3 {2(√(x−1))−2x}dx = 2 [(2/3)(x−1)^(3/2) ]_1 ^3 − [x^2 ]_1 ^3 =(4/3){ 2^(3/2) } −(9−1) = (4/3)(√2^3 ) −8 =(4/3)(2(√2)) −8 =((8(√2))/3) −8 .$
$3) \int_{1}^{3} f (x) d x = \int_{1}^{3} {2 \sqrt{x - 1} - 2 x} d x = 2 {[\frac{2}{3} {(x - 1)}^{\frac{3}{2}}]}_{1}^{3} - {[x^{2}]}_{1}^{3}$ $= \frac{4}{3} {2^{\frac{3}{2}}} - (9 - 1) = \frac{4}{3} \sqrt{2^{3}} - 8 = \frac{4}{3} (2 \sqrt{2}) - 8 = \frac{8 \sqrt{2}}{3} - 8 .$
Commented by maxmathsup by imad last updated on 31/Aug/18

$4) f (x) = y \Leftrightarrow x = f^{- 1} (y) a n d f (x) = y \Rightarrow 2 \sqrt{x - 1} - 2 x = y (x ⩾ 1) \Rightarrow$ $2 \sqrt{x - 1} = 2 x + y \Rightarrow 4 (x - 1) = {(2 x + y)}^{2} \Rightarrow 4 x - 4 = 4 x^{2} + 4 x y + y^{2} \Rightarrow$ $4 x^{2} + 4 x y + y^{2} - 4 x + 4 = 0 \Rightarrow 4 x^{2} + 4 (y - 1) x + y^{2} + 4 = 0$ $Δ^{^{'}} = 4 {(y - 1)}^{2} - 4 (y^{2} + 4) = 4 y^{2} - 8 y + 4 - 4 y^{2} - 16 = - 8 y - 12$ $t h i s e q u a t i o n h a v e s o l u t i o n \Leftrightarrow - 8 y - 12 ⩾ 0 \Leftrightarrow 8 y + 12 ⩽ 0 \Leftrightarrow y ⩽ \frac{- 12}{8} (= - \frac{3}{2})$ $x_{1} = \frac{- 2 (y - 1) + \sqrt{- 8 y - 12}}{4} = \frac{- (y - 1) + \sqrt{- 2 y - 3}}{2}$ $x_{2} = \frac{- (y - 1) - \sqrt{- 2 y - 3}}{2}$ $x_{1} - 1 = \frac{1 - y + \sqrt{- 2 y - 3}}{2} - 1 = \frac{- 1 - y + \sqrt{- 2 y - 3}}{2}$ $b u t y ⩽ - \frac{3}{2} \Rightarrow - y ⩾ \frac{3}{2} \Rightarrow - 1 - y ⩾ \frac{1}{2} ⩾ 0 \Rightarrow x_{1} - 1 ⩾ 0 n o n e e d t o v e r i f y x_{2}$ $\Rightarrow f^{- 1} (y) = \frac{1 - y + \sqrt{- 2 y - 3}}{2} \Rightarrow f^{- 1} (x) = \frac{1 - x + \sqrt{- 2 x - 3}}{2}$ $w i t h x ⩽ - \frac{3}{2} .$
Commented by maxmathsup by imad last updated on 31/Aug/18

$4) f o r g i v e t h e Q i s c a l c u l a t e \int_{- 3}^{- 2} f^{- 1} (x) d x$ $5) t h e Q i s f i n d t h e v a l u e o f A = \frac{\int_{2}^{3} f (x) d x}{\int_{- 3}^{- 2} f^{- 1} (x) d x}$ $B = \int \frac{f (x)}{f^{- 1} (x)} d x .$
Commented by maxmathsup by imad last updated on 31/Aug/18
$4) we have f^(−1) (x)=((1−x+(√(−2x−3)))/2) ⇒ ∫_(−3) ^(−2) f^(−1) (x)dx =(1/2) ∫_(−3) ^(−2) (1−x) +(1/2) ∫_(−3) ^(−2) (√(−2x−3))dx but ∫_(−3) ^(−2) (1−x)dx =[x−(x^2 /2)]_(−3) ^(−2) =−2−2−(−3−(9/2))=−1 +(9/2) =(7/2) and ∫_(−3) ^(−2) (√(−2x−3))dx =[−(1/3)(−2x−3)^(3/2) ]_(−3) ^(−2) =−(1/3){1−(3)^(3/2) } =(√3) −(1/3) ⇒∫_(−3) ^(−2) f^(−1) (x)dx = (7/4) +((√3)/2) −(1/6) =((38)/(24)) +((√3)/2) =((19)/(12)) +((√3)/2)$
$4) w e h a v e f^{- 1} (x) = \frac{1 - x + \sqrt{- 2 x - 3}}{2} \Rightarrow$ $\int_{- 3}^{- 2} f^{- 1} (x) d x = \frac{1}{2} \int_{- 3}^{- 2} (1 - x) + \frac{1}{2} \int_{- 3}^{- 2} \sqrt{- 2 x - 3} d x b u t$ $\int_{- 3}^{- 2} (1 - x) d x = {[x - \frac{x^{2}}{2}]}_{- 3}^{- 2} = - 2 - 2 - (- 3 - \frac{9}{2}) = - 1 + \frac{9}{2} = \frac{7}{2} a n d$ $\int_{- 3}^{- 2} \sqrt{- 2 x - 3} d x = {[- \frac{1}{3} {(- 2 x - 3)}^{\frac{3}{2}}]}_{- 3}^{- 2} = - \frac{1}{3} {1 - {(3)}^{\frac{3}{2}}}$ $= \sqrt{3} - \frac{1}{3} \Rightarrow \int_{- 3}^{- 2} f^{- 1} (x) d x = \frac{7}{4} + \frac{\sqrt{3}}{2} - \frac{1}{6} = \frac{38}{24} + \frac{\sqrt{3}}{2} = \frac{19}{12} + \frac{\sqrt{3}}{2}$
Commented by maxmathsup by imad last updated on 31/Aug/18
$5) we have ∫_2 ^3 f(x)dx = 2 ∫_2 ^3 (√(x−1)) −∫_2 ^3 2x dx =2[(2/3)(x−1)^(3/2) ]_2 ^3 −[x^2 ]_2 ^3 =(4/3){ 2^(3/2) −1}−{9−4} =(4/3)(2(√2)) −5 =((8(√2))/3) −5 also we have ∫_(−3) ^(−2) f^(−1) (x)dx =((19)/(12)) +((√3)/2) ⇒ A = ((∫_2 ^3 f(x)dx)/(∫_(−3) ^(−2) f^(−1) (x)dx)) = ((((8(√2))/3)−5)/(((19)/(12)) +((√3)/2))) .$
$5) w e h a v e \int_{2}^{3} f (x) d x = 2 \int_{2}^{3} \sqrt{x - 1} - \int_{2}^{3} 2 x d x$ $= 2 {[\frac{2}{3} {(x - 1)}^{\frac{3}{2}}]}_{2}^{3} - {[x^{2}]}_{2}^{3} = \frac{4}{3} {2^{\frac{3}{2}} - 1} - {9 - 4}$ $= \frac{4}{3} (2 \sqrt{2}) - 5 = \frac{8 \sqrt{2}}{3} - 5 a l s o w e h a v e \int_{- 3}^{- 2} f^{- 1} (x) d x = \frac{19}{12} + \frac{\sqrt{3}}{2} \Rightarrow$ $A = \frac{\int_{2}^{3} f (x) d x}{\int_{- 3}^{- 2} f^{- 1} (x) d x} = \frac{\frac{8 \sqrt{2}}{3} - 5}{\frac{19}{12} + \frac{\sqrt{3}}{2}} .$
	Answered by behi83417@gmail.com last updated on 30/Aug/18

	$1) x - 1 ⩾ 0 \Rightarrow x ⩾ 1 \Rightarrow D_{f} = [1, + \infty)$ $2) y = 2 \sqrt{x - 1} - 2 x$ $y^{'} = \frac{1}{\sqrt{x - 1}} - 2 \Rightarrow x \neq 1 \Rightarrow D_{f} = (1, + \infty)$ $x - 1 = t^{2} \Rightarrow x = t^{2} + 1$ $f (t) = 2 t - 2 (t^{2} + 1) = - 2 (t^{2} - t + 1)$ $\Rightarrow t^{2} - t + 1 = \frac{- y}{2} \Rightarrow {(t - \frac{1}{2})}^{2} = \frac{- y}{2} - \frac{3}{4}$ $\Rightarrow t - \frac{1}{2} = \pm \sqrt{- \frac{2 y + 3}{4}} \Rightarrow t = \frac{1}{2} (1 \pm \sqrt{- 2 y - 3})$ $\Rightarrow \sqrt{x - 1} = \frac{1}{2} (1 \pm \sqrt{- 2 y - 3})$ $\Rightarrow x = 1 + \frac{1}{4} {(1 \pm \sqrt{- 2 y - 3})}^{2}$ $\Rightarrow f^{- 1} (x) = 1 + \frac{1}{4} {(1 \pm \sqrt{- 2 x - 3})}^{2}$ $- 2 y - 3 ⩾ 0 \Rightarrow 2 y + 3 ⩽ 0 \Rightarrow y ⩽ - \frac{3}{2}$ $\Rightarrow R_{f} = (- \infty, - \frac{3}{2}] .$ $3) \underset{1}{\int^{3}} f (x) d x = \underset{1}{\int^{3}} (2 \sqrt{x - 1} - 2 x) d x =$ $Missing \left or extra \right$ $4) \int f^{- 1} (x) d x = \int [1 + \frac{1}{4} {(1 \pm \sqrt{- 2 x - 3})}^{2}] d x =$ $= \int [1 + \frac{1}{4} (1 \mp (2 x + 3) \pm 2 \sqrt{- 2 x - 3})] d x$ $I_{1} = \int [1 + \frac{1}{4} (2 x + 4 + 2 \sqrt{- 2 x - 3})] d x =$ $Missing \left or extra \right$ $= \frac{1}{2} [6 + 8 - 27 i - 2 - 2 + \frac{i \sqrt{5}}{3}] =$ $= 5 - \frac{81 - \sqrt{5}}{6} . i$
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