let f(x)=ln(√((2+x)/(2−x))) 1) find D_f and find the assymptotes to C_f 2) calculate f^′ (x) and give the variation of f 3) give the graph of f 4) give the equation of tangent to C_(f ) at point E((1/2),f((1/2))) 5) calculate ∫


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Question Number 40097 by maxmathsup by imad last updated on 15/Jul/18

$l e t f (x) = l n \sqrt{\frac{2 + x}{2 - x}}$ $1) f i n d D_{f} a n d f i n d t h e a s s y m p t o t e s t o C_{f}$ $2) c a l c u l a t e f^{^{'}} (x) a n d g i v e t h e v a r i a t i o n o f f$ $3) g i v e t h e g r a p h o f f$ $4) g i v e t h e e q u a t i o n o f t a n g e n t t o C_{f} a t p o i n t E (\frac{1}{2}, f (\frac{1}{2}))$ $5) c a l c u l a t e \int_{0}^{1} f (x) d x .$
Commented by math khazana by abdo last updated on 03/Aug/18
$1) D_f =]−2,2[ lim_(x→−2^+ ) f(x)=−∞ lim_(x→2^− ) f(x)=+∞ so x=−2 and x=2 are assymptotes to C_f 2) we have f(x)=(1/2){ln(2+x)−ln(2−x)}⇒ f^′ (x)= (1/(2(2+x))) +(1/(2(2−x))) =(1/2){(1/(2+x)) +(1/(2−x))} =(1/2) (4/(4−x^2 )) =(2/(4−x^2 )) > 0 because −2<x<2 so f is increasing on]−2,2[ 4) equ.of tangent is y=f^′ ((1/2))(x−(1/2))+f((1/2)) f^′ ((1/2))= (2/(4−(1/4))) =((2×4)/(15)) =(8/(15)) f((1/2)) =(1/2)ln(((2+(1/2))/(2−(1/2)))) =(1/2)ln((5/3)) ⇒ y =(8/(15))(x−(1/2)) +(1/2)ln((5/3)) y=(8/(15))x −(4/(15)) +(1/2)ln((5/3)).$
$1) D_{f} =] - 2, 2 [$ ${l i m}_{x \to - 2^{+}} f (x) = - \infty$ ${l i m}_{x \to 2^{-}} f (x) = + \infty s o x = - 2 a n d x = 2 a r e$ $a s s y m p t o t e s t o C_{f}$ $2) w e h a v e f (x) = \frac{1}{2} {l n (2 + x) - l n (2 - x)} \Rightarrow$ $f^{^{'}} (x) = \frac{1}{2 (2 + x)} + \frac{1}{2 (2 - x)} = \frac{1}{2} {\frac{1}{2 + x} + \frac{1}{2 - x}}$ $= \frac{1}{2} \frac{4}{4 - x^{2}} = \frac{2}{4 - x^{2}} > 0 b e c a u s e - 2 < x < 2 s o f$ $i s i n c r e a s i n g o n] - 2, 2 [$ $4) e q u . o f t a n g e n t i s y = f^{^{'}} (\frac{1}{2}) (x - \frac{1}{2}) + f (\frac{1}{2})$ $f^{^{'}} (\frac{1}{2}) = \frac{2}{4 - \frac{1}{4}} = \frac{2 \times 4}{15} = \frac{8}{15}$ $f (\frac{1}{2}) = \frac{1}{2} l n (\frac{2 + \frac{1}{2}}{2 - \frac{1}{2}}) = \frac{1}{2} l n (\frac{5}{3}) \Rightarrow$ $y = \frac{8}{15} (x - \frac{1}{2}) + \frac{1}{2} l n (\frac{5}{3})$ $y = \frac{8}{15} x - \frac{4}{15} + \frac{1}{2} l n (\frac{5}{3}) .$
Commented by math khazana by abdo last updated on 03/Aug/18

$5) l e t I = \int_{0}^{1} l n (\sqrt{\frac{2 + x}{2 - x}}) d x$ $I = \frac{1}{2} \int_{0}^{1} l n (2 + x) d x - \frac{1}{2} \int_{0}^{1} l n (2 - x) d x b u t$ $\int_{0}^{1} l n (2 + x) d x =_{2 + x = t} \int_{2}^{3} l n (t) d t = {[t l n (t) - t]}_{2}^{3}$ $= 3 l n (3) - 3 - 2 l n (2) + 2 = 3 l n (3) - 2 l n (2) - 1$ $\int_{0}^{1} l n (2 - x) d x =_{2 - x = t} - \int_{2}^{1} l n (t) d t$ $= \int_{1}^{2} l n (t) d t = {[t l n (t) - t]}_{1}^{2} = 2 l n (2) - 2 + 1$ $= 2 l n (2) - 1 \Rightarrow$ $\int_{0}^{1} f (x) d x = \frac{3}{2} l n (3) - l n (2) - \frac{1}{2} - l n (2) + \frac{1}{2}$ $= \frac{3}{2} l n (3) - 2 l n (2) .$
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