let f(x)= (√(2+x^2 )) −x 1) calculate lim_(x→+∞) f(x) and lim_(x→−∞) f(x) 2) calculate lim_(x→+∞) ((f(x))/x) and lim_(x→−∞) ((f(x))/x) 3) calculate f^′ (x) and determine its sign 4) give the variation of 5) give the equation of assymptote at point A(1,f(1)) 6) find f^(−1) (x) and calculate (f^(−1) )^′ (x) 7) calculate ∫


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 36442 by abdo mathsup 649 cc last updated on 02/Jun/18

$l e t f (x) = \sqrt{2 + x^{2}} - x$ $1) c a l c u l a t e {l i m}_{x \to + \infty} f (x) a n d {l i m}_{x \to - \infty} f (x)$ $2) c a l c u l a t e {l i m}_{x \to + \infty} \frac{f (x)}{x} a n d {l i m}_{x \to - \infty} \frac{f (x)}{x}$ $3) c a l c u l a t e f^{^{'}} (x) a n d d e t e r m i n e i t s s i g n$ $4) g i v e t h e v a r i a t i o n o f$ $5) g i v e t h e e q u a t i o n o f a s s y m p t o t e a t p o i n t$ $A (1, f (1))$ $6) f i n d f^{- 1} (x) a n d c a l c u l a t e {(f^{- 1})}^{^{'}} (x)$ $7) c a l c u l a t e \int_{0}^{4} f (x) d x .$
Commented by abdo.msup.com last updated on 05/Jun/18

$1) {l i m}_{x \to + \infty} f (x) = {l i m}_{x \to + \infty} \frac{2}{\sqrt{2 + x^{2}} + x}$ $= 0 a n d {l i m}_{x \to - \infty} f (x) = + \infty$ $2) {l i m}_{x \to + \infty} \frac{f (x)}{x} = l i m \frac{\sqrt{2 + x^{2}} - x}{x}$ $= {l i m}_{x \to + \infty} \sqrt{\frac{2}{x^{2}} + 1} - 1 = 0$ ${l i m}_{x \to - \infty} \frac{f (x)}{x} = {l i m}_{x \to - \infty} \frac{\sqrt{2 + x^{2}} - x}{x}$ $= {l i m}_{x \to - \infty} - \sqrt{\frac{2}{x^{2}} + 1} - 1 = - 2$ $3) w e h a v e f (x) = \sqrt{2 + x^{2}} - x \Rightarrow$ $f^{^{'}} (x) = \frac{2 x}{2 \sqrt{2 + x^{2}}} - 1 = \frac{x}{\sqrt{2 + x^{2}}} - 1$ $= \frac{x - \sqrt{2 + x^{2}}}{\sqrt{2 + x^{2}}} i f x ⩽ 0 f^{^{'}} (x) ⩽ 0$ $i f x ⩾ 0 w e h a v e x^{2} - {(\sqrt{2 + x^{2}})}^{2} ⩽ 0 \Rightarrow$ $f^{^{'}} (x) ⩽ 0$ $4) f i s d e c r e a s i n g o n R .$
Commented by abdo.msup.com last updated on 05/Jun/18

$5) w e h a v e f (x) = \sqrt{2 + x^{2}} - x \Rightarrow$ $f (1) = \sqrt{3} - 1 a l s o f^{^{'}} (x) = \frac{x}{\sqrt{2 + x^{2}}} - 1 \Rightarrow$ $f^{^{'}} (1) = \frac{1}{\sqrt{3}} - 1 s o t h e e q u s t i o n o f$ $a s s y m p t o t e i s$ $y = f^{^{'}} (1) (x - 1) + f (1) \Rightarrow$ $y = (\frac{1}{\sqrt{3}} - 1) (x - 1) + \sqrt{3} - 1 .$
Commented by abdo.msup.com last updated on 05/Jun/18

$6) f i s d e c r e a s i n g s o f] - \infty, + \infty [$ $=] f (+ \infty), f (- \infty) [=] 0, + \infty [f i s$ $a b i j e c t i o n f r o m R t o] 0, + \infty [$ $l e t f (x) = y \Leftrightarrow \sqrt{2 + x^{2}} - x = y \Rightarrow$ $\sqrt{2 + x^{2}} = x + y (s o x + y > 0) \Rightarrow$ $2 + x^{2} = x^{2} + 2 x y + y^{2} \Rightarrow$ $2 = 2 x y + y^{2} \Rightarrow 2 - y^{2} = 2 x y \Rightarrow x = \frac{2 - y^{2}}{2 y}$ $\Rightarrow f^{- 1} (x) = \frac{2 - x^{2}}{2 x} a n d x > 0$
Commented by abdo.msup.com last updated on 05/Jun/18

$w e h a v e f^{- 1} (x) = \frac{1}{x} - \frac{x}{2} \Rightarrow$ ${(f^{- 1} (x))}^{^{'}} = - \frac{1}{x^{2}} - \frac{1}{2} .$
Commented by abdo.msup.com last updated on 05/Jun/18
$7) let A= ∫_0 ^4 f(x)dx A = ∫_0 ^4 ((√(2+x^2 )) −x)dx =∫_0 ^4 (√(2+x^2 )) dx −[(x^2 /2)]_0 ^4 =∫_0 ^4 (√(2+x^2 ))dx −8 but chang.x=(√2) sh(t) give ∫_0 ^4 (√(2+x^2 ))dx=(√2) ∫_0 ^(argsh((4/(√2)))) ch(t)(√2)chtdt = 2 ∫_0 ^(argsh((4/(√2)))) ch^2 (t)dt =2 ∫_0 ^(argsh((4/(√2)))) ((1+ch(2t))/2)dt =argsh((4/(√2))) +(1/2)[sh(2t)]_0 ^(argsh((4/(√2)))) =ln( (4/(√2)) +3) +(1/2)sh(2 ln{(4/(√2)) +3)}) =ln((4/(√2)) +3) +(1/4){ ((4/(√2)) +3)^2 +((4/(√2)) +3)^(−1) } .$
$7) l e t A = \int_{0}^{4} f (x) d x$ $A = \int_{0}^{4} (\sqrt{2 + x^{2}} - x) d x$ $= \int_{0}^{4} \sqrt{2 + x^{2}} d x - {[\frac{x^{2}}{2}]}_{0}^{4}$ $= \int_{0}^{4} \sqrt{2 + x^{2}} d x - 8 b u t c h a n g . x = \sqrt{2} s h (t)$ $g i v e$ $\int_{0}^{4} \sqrt{2 + x^{2}} d x = \sqrt{2} \int_{0}^{a r g s h (\frac{4}{\sqrt{2}})} c h (t) \sqrt{2} c h t d t$ $= 2 \int_{0}^{a r g s h (\frac{4}{\sqrt{2}})} {c h}^{2} (t) d t = 2 \int_{0}^{a r g s h (\frac{4}{\sqrt{2}})} \frac{1 + c h (2 t)}{2} d t$ $= a r g s h (\frac{4}{\sqrt{2}}) + \frac{1}{2} {[s h (2 t)]}_{0}^{a r g s h (\frac{4}{\sqrt{2}})}$ $= l n (\frac{4}{\sqrt{2}} + 3) + \frac{1}{2} s h (2 l n {\frac{4}{\sqrt{2}} + 3)})$ $= l n (\frac{4}{\sqrt{2}} + 3) + \frac{1}{4} {{(\frac{4}{\sqrt{2}} + 3)}^{2} + {(\frac{4}{\sqrt{2}} + 3)}^{- 1}} .$
Commented by abdo.msup.com last updated on 05/Jun/18
$I =ln(4+3(√2)) −(1/2)ln(2) +(1/4){ ((4/(√2)) +3)^2 −((4/(√2)) +3)^(−2) } −8 .$
$I = l n (4 + 3 \sqrt{2}) - \frac{1}{2} l n (2)$ $+ \frac{1}{4} {{(\frac{4}{\sqrt{2}} + 3)}^{2} - {(\frac{4}{\sqrt{2}} + 3)}^{- 2}} - 8 .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum