1) study the function f(x)=ln(x+1−(√x)) 2) determine f^(−1) (x) 3) cslculate ∫ f(x)dx snd ∫ f^(−1) (x)dx 4) dtermine ∫ f^(−1) (x^2 +f(x))dx


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Question Number 53623 by turbo msup by abdo last updated on 24/Jan/19

$1) s t u d y t h e f u n c t i o n$ $f (x) = l n (x + 1 - \sqrt{x})$ $2) d e t e r m i n e f^{- 1} (x)$ $3) c s l c u l a t e \int f (x) d x s n d$ $\int f^{- 1} (x) d x$ $4) d t e r m i n e \int f^{- 1} (x^{2} + f (x)) d x$
Commented by Abdo msup. last updated on 25/Jan/19

$1) x \in D_{f} \Leftrightarrow x + 1 - \sqrt{x} > 0 \Leftrightarrow t^{2} + 1 - t > 0 (t = \sqrt{x})$ $t^{2} - t + 1 > 0 \to Δ = 1 - 4 = - 3 < 0 a = 1 > 0 \Rightarrow$ $t^{2} - t + 1 > 0 \forall t \Rightarrow D_{f} = [0, + \infty [$ $w e h a v e f (x) = l n (x) + l n (1 + \frac{1}{x} - \frac{1}{\sqrt{x}}) \Rightarrow$ ${l i m}_{x \to + \infty} f (x) = {l i m}_{x \to + \infty} l n (x) = + \infty$ ${l i m}_{x \to + \infty} \frac{f (x)}{x} = {l i m}_{x \to + \infty} \frac{l n (x)}{x} = 0 s o t h e g r a p h$ $h a v e a p a r a b o l b r a n c h i n (o y) d i r e c t i o n$ $d e r i v a t i v e f^{^{'}} (x) = \frac{{(x + 1 - \sqrt{x})}^{,}}{x + 1 - \sqrt{x}}$ $= \frac{1 - \frac{1}{2 \sqrt{x}}}{x + 1 - \sqrt{x}} = \frac{2 \sqrt{x} - 1}{2 \sqrt{x} (x + 1 - \sqrt{x})}$ $f^{^{'}} (x) = 0 \Leftrightarrow 2 \sqrt{x} - 1 = 0 \Leftrightarrow \sqrt{x} = \frac{1}{2} \Leftrightarrow x = \frac{1}{4}$ $f^{^{'}} (x) ⩾ 0 \Leftrightarrow x ⩾ \frac{1}{4}$ $v a r i a t i o n s o f f (x)$ $- - - - - - - - - - - - - - - - -$ $x 0 \frac{1}{4} + \infty$ $f^{,} (x) - 0 +$ $f (x) 0 d e c r . f (\frac{1}{4}) i n c r .$ $- - - - - - - - - - - - - - - - - - -$ $f (\frac{1}{4}) = l n (\frac{5}{4} - \frac{1}{2}) = l n (\frac{3}{4}) = l n (3) - 2 l n (2)$ $i n t e r s e c t i o n b e t w e e n g r a p h a n d (o x)$ $f (x) = 0 \Leftrightarrow l n (x + 1 - \sqrt{x}) = 0 \Leftrightarrow x + 1 - \sqrt{x} = 1$ $x = \sqrt{x} \Leftrightarrow x^{2} = x \Leftrightarrow x = 0 o r x = 1$ $t h e p o i n t s a r e 0 (0, 0) a n d A (1, 0)$ $r e s t d r a w i n g g r a p h . . b e c i n t i n u e d . . .$
Commented by Abdo msup. last updated on 25/Jan/19

$2) f (x) = y \Leftrightarrow x = f^{- 1} (y) w i t h x ⩾ 0$ $f (x) = y \Leftrightarrow l n (x + 1 - \sqrt{x}) = y \Leftrightarrow x + 1 - \sqrt{x} = e^{y} \Leftrightarrow$ $x + 1 - \sqrt{x} - e^{y} = 0 \Rightarrow t^{2} + 1 - t - e^{y} = 0 w i t h t = \sqrt{x} \Rightarrow$ $t^{2} - t + 1 - e^{y} = 0 \to Δ = 1 - 4 (1 - e^{y})$ $= 4 e^{y} - 3 f^{- 1} e x i s t s \Rightarrow Δ ⩾ 0 \Rightarrow e^{y} ⩾ \frac{3}{4} \Rightarrow y ⩾ l n (\frac{3}{4})$ $i n t h i s c a s e t_{1} = \frac{1 + \sqrt{4 e^{y} - 3}}{2}$ $a n d t_{2} = \frac{1 - \sqrt{4 e^{y} - 3}}{2} b u t t ⩾ 0 \Rightarrow t = t_{1} = \sqrt{x} \Rightarrow$ $x = t_{1}^{2} = \frac{1}{4} {(1 + \sqrt{4 e^{y} - 3})}^{2} \Rightarrow$ $f^{- 1} (x) = \frac{1}{4} {(1 + \sqrt{4 e^{x} - 3})}^{2}$
Commented by Abdo msup. last updated on 25/Jan/19

$3) l e t I = \int f (x) d x \Rightarrow I = \int l n (x + 1 - \sqrt{x}) d x$ $=_{\sqrt{x} = t} \int l n (t^{2} + 1 - t) (2 t) d t$ $= 2 \int t l n (t^{2} - t + 1) d t b y p a r t s$ $\int t l n (t^{2} - t + 1) d t = \frac{t^{2}}{2} l n (t^{2} - t + 1) - \int \frac{t^{2}}{2} \frac{2 t - 1}{t^{2} - t + 1} d t$ $= \frac{t^{2}}{2} l n (t^{2} - t + 1) - \frac{1}{2} \int \frac{2 t^{3} - t^{2}}{t^{2} - t + 1} d t$ $\int \frac{2 t^{3} - t^{2}}{t^{2} - t + 1} d t = \int \frac{2 t (t^{2} - t + 1) + 2 t^{2} - 2 t - t^{2}}{t^{2} - t + 1} d t$ $= \int (2 t) d t + \int \frac{t^{2} - 2 t}{t^{2} - t + 1} d t$ $= t^{2} + \int \frac{t^{2} - t + 1 - t - 1}{t^{2} - t + 1} d t$ $= t^{2} + t - \int \frac{t + 1}{t^{2} - t + 1} d t$ $\int \frac{t + 1}{t^{2} - t + 1} d t = \frac{1}{2} \int \frac{2 t + 2}{t^{2} - t + 1} d t$ $= \frac{1}{2} \int \frac{2 t - 1 + 3}{t^{2} - t + 1} d t = \frac{1}{2} l n (t^{2} - t + 1 + \frac{3}{2} \int \frac{d t}{t^{2} - t + 1}$ $= . . . . b e c o n t i n u e d . . .$
	Answered by kaivan.ahmadi last updated on 24/Jan/19
	$first we find D_f x+1−(√x) >0⇒x+1>(√x)⇒x^2 +2x+1>x⇒ x^2 +x+1>0 since x^2 +x+1=0 has no real root and the coefficient of x^2 is positive, then x∈R on the other hand D_(√x) ={x∈R∣x≥0}. so D_f ={x∈R∣x≥0} now we check 1−1 (1,0) , (0,0) ∈f so f is not 1−1 consequently f^(−1) is not exist$
	$first we find D_{f}$ $x + 1 - \sqrt{x} > 0 \Rightarrow x + 1 > \sqrt{x} \Rightarrow x^{2} + 2 x + 1 > x \Rightarrow$ $x^{2} + x + 1 > 0$ $since x^{2} + x + 1 = 0 has no real root and$ $the coefficient of x^{2} is positive, then x \in R$ $on the other hand D_{\sqrt{x}} = {x \in R ∣ x ⩾ 0} .$ $so D_{f} = {x \in R ∣ x ⩾ 0}$ $now we check 1 - 1$ $(1, 0), (0, 0) \in f so f is not 1 - 1$ $consequently f^{- 1} is not exist$
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