let f(x)=ln(x+1−2(√x)) 1) find D_f 2) determine f^(−1) 3) calculate ∫f (x)dx and ∫ f^(−1) (x)dx


Question and Answers Forum

All Questions Topic List
Relation and Functions Questions
Previous in All Question Next in All Question
Previous in Relation and Functions Next in Relation and Functions
Question Number 62129 by maxmathsup by imad last updated on 15/Jun/19

$l e t f (x) = l n (x + 1 - 2 \sqrt{x})$ $1) f i n d D_{f}$ $2) d e t e r m i n e f^{- 1}$ $3) c a l c u l a t e \int f (x) d x a n d \int f^{- 1} (x) d x$
Commented by arcana last updated on 16/Jun/19
$1) x+1−2(√x)>0⇒((√x)−1)^2 >0 necessary (√x)−1>0 ∨ (√x)−1<0 (√x)>1∨(√x)<1 x>1∨x<1, x≥0 ⇒D_f =[0,1)∪(1,+∞)=R^+ −{1}$
$1)$ $x + 1 - 2 \sqrt{x} > 0 \Rightarrow {(\sqrt{x} - 1)}^{2} > 0$ $n e c e s s a r y \sqrt{x} - 1 > 0 \lor \sqrt{x} - 1 < 0$ $\sqrt{x} > 1 \lor \sqrt{x} < 1$ $x > 1 \lor x < 1, x ⩾ 0$ $\Rightarrow D_{f} = [0, 1) \cup (1, + \infty) = R^{+} - {1}$
Commented by kaivan.ahmadi last updated on 16/Jun/19

$e^{y} = {(\sqrt{x} - 1)}^{2} \Rightarrow \sqrt{e^{y}} = \sqrt{x} - 1 \Rightarrow 1 + \sqrt{e^{y}} = \sqrt{x} \Rightarrow$ $x = \sqrt{1 + \sqrt{e^{y}}} \Rightarrow f^{- 1} (x) = \sqrt{1 + \sqrt{e^{x}}}$
Commented by maxmathsup by imad last updated on 16/Jun/19

$1) x \in D_{f} \Leftrightarrow x ⩾ 0 a n d x - 2 \sqrt{x} + 1 > 0 \Leftrightarrow x ⩾ 0 a n d {(\sqrt{x} - 1)}^{2} > 0 \Leftrightarrow x ⩾ 0 a n d x \neq 1 \Rightarrow$ $D_{f} = [0, 1 [\cup] 1, + \infty [$ $2) f (x) = y \Leftrightarrow x = f^{- 1} (y)$ $f (x) = y \Rightarrow l n (x + 1 - 2 \sqrt{x}) = y \Rightarrow x + 1 - 2 \sqrt{x} = e^{y} \Rightarrow {(\sqrt{x} - 1)}^{2} = e^{y} \Rightarrow$ $∣ \sqrt{x} - 1 ∣ = \sqrt{e^{y}} b y e^{y} > 0 \Rightarrow x > 1 \Rightarrow \sqrt{x} - 1 = \sqrt{e^{y}} \Rightarrow \sqrt{x} = 1 + e^{\frac{y}{2}} \Rightarrow$ $x = {(1 + e^{\frac{y}{2}})}^{2} \Rightarrow f^{- 1} (x) = {(1 + e^{\frac{x}{2}})}^{2}$ $3) l e t I = \int f (x) d x \Rightarrow I = \int l n (x + 1 - 2 \sqrt{x}) d x b y p a r t s$ $I = x l n (x + 1 - 2 \sqrt{x}) - \int x \frac{1 - \frac{1}{\sqrt{x}}}{x + 1 - 2 \sqrt{x}} d x$ $= x l n (x + 1 - 2 \sqrt{x}) - \int x \frac{\sqrt{x} - 1}{\sqrt{x} (x + 1 - 2 \sqrt{x})} d x$ $\int x \frac{\sqrt{x} - 1}{\sqrt{x} (x + 1 - 2 \sqrt{x})} d x = \int \frac{\sqrt{x} (\sqrt{x} - 1)}{x + 1 - 2 \sqrt{x}} d x = \int \frac{\sqrt{x} (\sqrt{x} - 1)}{{(\sqrt{x} - 1)}^{2}} d x = \int \frac{\sqrt{x}}{\sqrt{x} - 1} d x$ $=_{\sqrt{x} = t} \int \frac{t}{t - 1} (2 t) d t = 2 \int \frac{t^{2} - 1 + 1}{t - 1} d t = 2 \int (t + 1) d t + 2 \int \frac{d t}{t - 1}$ $= 2 \frac{t^{2}}{2} + 2 t + 2 l n ∣ t - 1 ∣ + c = t^{2} + 2 t + 2 l n ∣ t - 1 ∣ + c$ $= x + 2 \sqrt{x} + 2 l n ∣ \sqrt{x} - 1 ∣ + c \Rightarrow$ $I = x l n (x + 1 - 2 \sqrt{x}) - x - 2 \sqrt{x} - 2 l n ∣ \sqrt{x} - 1 ∣ + c .$
Commented by maxmathsup by imad last updated on 16/Jun/19

$\int f^{- 1} (x) d x = \int {(1 + e^{\frac{x}{2}})}^{2} d x = \int (1 + 2 e^{\frac{x}{2}} + e^{x}) d x$ $= x + 4 e^{\frac{x}{2}} + e^{x} + c .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum