let f(x)=(√(1+2x^2 )) −x(√2) +3 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)give the assymtote to graph C_f 4) give the assymtote to C_f at point A(0,f(0)) 5) find f^(−1) (x) and calculate (f^(−1) )^′ (x) 6) calculate ∫


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Question Number 38721 by maxmathsup by imad last updated on 28/Jun/18

$l e t f (x) = \sqrt{1 + 2 x^{2}} - x \sqrt{2} + 3$ $1) c a l c u l a t e {l i m}_{x \to + \infty} f (x) a n d {l i m}_{x \to - \infty} f (x)$ $2) c a l c u l a t e {l i m}_{x \to + \infty} \frac{f (x)}{x} a n d {l i m}_{x \to - \infty} \frac{f (x)}{x}$ $3) g i v e t h e a s s y m t o t e t o g r a p h C_{f}$ $4) g i v e t h e a s s y m t o t e t o C_{f} a t p o i n t A (0, f (0))$ $5) f i n d f^{- 1} (x) a n d c a l c u l a t e {(f^{- 1})}^{^{'}} (x)$ $6) c a l c u l a t e \int_{0}^{1} f (x) d x .$
Commented by prof Abdo imad last updated on 02/Jul/18
$1) we have (√(1+2x^2 )) =x(√2)(√( 1+(1/(2x^2 )))) ∼ x(√2){ 1+ (1/(4x^2 ))} (x→+∞) ⇒f(x)∼ ((√2)/(4x)) +3 ⇒ lim_(x→+∞) f(x)= 3 also we can use this method lim_(x→+∞) f(x)=lim((((√(1+2x^2 ))−(x(√2)−3))((√(1+2x^2 )) +x(√2)−3))/((√(1+2x^2 )) +x(√2) −3)) =lim_(x→+∞) ((1+2x^2 −(2x^2 −6x(√2) +9))/((√(1+2x^2 )) +x(√2) −3)) =lim_(x→+∞) ((6x(√2) −8)/((√(1+2x^2 )) +x(√2)−3)) =lim_(x→+∞) ((6x(√2)−8)/(x(√2){(√(1+(1/(2x^2 )))) +1−(3/(x(√2)))})) =3 lim_(x→−∞) f(x)=lim_(x→−∞) (√(1+2x^2 )) −x(√2) +3 =+∞ 2)we have lim_(x→+∗) f(x)=3 ⇒lim_(x→+∞) ((f(x))/x) =0 (√(1+2x^2 ))=−x(√2)(√(1+(1/(2x^2 )))) for x<0 ∼−x(√2)( 1+(1/(4x^2 )))(x→−∞) ⇒ f(x) ∼ −2(√2)x −((√2)/(4x)) +3 ⇒ ((f(x))/x) ∼−2(√2) −((√2)/(4x^2 )) +(3/x) ⇒lim_(x→+∞) ((f(x))/x) ^ =−2(√2)$
$1) w e h a v e \sqrt{1 + 2 x^{2}} = x \sqrt{2} \sqrt{1 + \frac{1}{2 x^{2}}} \sim$ $x \sqrt{2} {1 + \frac{1}{4 x^{2}}} (x \to + \infty) \Rightarrow f (x) \sim \frac{\sqrt{2}}{4 x} + 3 \Rightarrow$ ${l i m}_{x \to + \infty} f (x) = 3 a l s o w e c a n u s e t h i s m e t h o d$ ${l i m}_{x \to + \infty} f (x) = l i m \frac{(\sqrt{1 + 2 x^{2}} - (x \sqrt{2} - 3)) (\sqrt{1 + 2 x^{2}} + x \sqrt{2} - 3)}{\sqrt{1 + 2 x^{2}} + x \sqrt{2} - 3}$ $= {l i m}_{x \to + \infty} \frac{1 + 2 x^{2} - (2 x^{2} - 6 x \sqrt{2} + 9)}{\sqrt{1 + 2 x^{2}} + x \sqrt{2} - 3}$ $= {l i m}_{x \to + \infty} \frac{6 x \sqrt{2} - 8}{\sqrt{1 + 2 x^{2}} + x \sqrt{2} - 3}$ $= {l i m}_{x \to + \infty} \frac{6 x \sqrt{2} - 8}{x \sqrt{2} {\sqrt{1 + \frac{1}{2 x^{2}}} + 1 - \frac{3}{x \sqrt{2}}}} = 3$ ${l i m}_{x \to - \infty} f (x) = {l i m}_{x \to - \infty} \sqrt{1 + 2 x^{2}} - x \sqrt{2} + 3 = + \infty$ $2) w e h a v e {l i m}_{x \to + *} f (x) = 3 \Rightarrow {l i m}_{x \to + \infty} \frac{f (x)}{x} = 0$ $\sqrt{1 + 2 x^{2}} = - x \sqrt{2} \sqrt{1 + \frac{1}{2 x^{2}}} f o r x < 0$ $\sim - x \sqrt{2} (1 + \frac{1}{4 x^{2}}) (x \to - \infty) \Rightarrow$ $f (x) \sim - 2 \sqrt{2} x - \frac{\sqrt{2}}{4 x} + 3 \Rightarrow \frac{f (x)}{x} \sim - 2 \sqrt{2} - \frac{\sqrt{2}}{4 x^{2}} + \frac{3}{x}$ $\Rightarrow {l i m}_{x \to + \infty} \frac{f (x)}{x} \overset{}{} = - 2 \sqrt{2}$
Commented by prof Abdo imad last updated on 02/Jul/18

${l i m}_{x \to - \infty} \frac{f (x)}{x} = - 2 \sqrt{2} b u t w e h a v e$ $f (x) \sim - 2 \sqrt{2} x + 3 - \frac{\sqrt{2}}{4 x} (x \to - \infty) \Rightarrow$ ${l i m}_{x \to - \infty} f (x) - (- 2 \sqrt{2} x + 3) = 0 s o$ $y = - 2 \sqrt{2} + 3 i s e q u a t i o n o f a s s h m p t o t e t o C_{f}$ $a t - \infty$
Commented by prof Abdo imad last updated on 02/Jul/18

$4) f i r s t c h a n g e a s s y m p t o t e b y e q u a t i o n o f$ $t a n g e n t$ $w e h a v e f^{^{'}} (x) = \frac{4 x}{2 \sqrt{1 + 2 x^{2}}} - \sqrt{2} = \frac{2 x}{\sqrt{1 + 2 x^{2}}} - \sqrt{2}$ $\Rightarrow f^{^{'}} (0) = - \sqrt{2}$ $f (0) = 4 \Rightarrow t h e e q u a t i o n o f a s s y m p t o t e i s$ $y = f^{^{'}} (0) x + f (0) = - \sqrt{2} x + 4$
Commented by prof Abdo imad last updated on 02/Jul/18
$5) f(x)=y ⇔ x =f^(−1) (y) ⇒(√(1+2x^2 ))−x(√2)+3=y ⇒(√(1+2x^2 ))=(x(√2)+y−3)⇒ 1+2x^2 =(x(√2)+y−3)^2 ⇒ 1+2x^2 =2x^2 +2x(√2)(y−3) +y^2 −6y +9 ⇒ 2(√2)(y−3)x +y^(2 ) −6y +8=0 ⇒ 2(√2)(y−3)x = −y^2 +6y −8 ⇒ x=((−y^2 +6y −8)/(2(√2)(y−3))) =f^(−1) (y) ⇒ f^(−1) (x) = ((x^(2 ) −6x +8)/(2(√2)(3−x))) . (f^(−1) (x))^′ =(1/(2(√2))){ (((2x−6)(3−x) +(x^2 −6x +8))/((3−x)^2 ))} =(1/(2(√2))){ ((6x −2x^2 −18 +6x +x^(2 ) −6x +8)/((3−x)^2 ))} =(1/(2(√2))){ ((−x^2 +6x −10)/((3−x)^2 ))}$
$5) f (x) = y \Leftrightarrow x = f^{- 1} (y) \Rightarrow \sqrt{1 + 2 x^{2}} - x \sqrt{2} + 3 = y$ $\Rightarrow \sqrt{1 + 2 x^{2}} = (x \sqrt{2} + y - 3) \Rightarrow$ $1 + 2 x^{2} = {(x \sqrt{2} + y - 3)}^{2} \Rightarrow$ $1 + 2 x^{2} = 2 x^{2} + 2 x \sqrt{2} (y - 3) + y^{2} - 6 y + 9 \Rightarrow$ $2 \sqrt{2} (y - 3) x + y^{2} - 6 y + 8 = 0 \Rightarrow$ $2 \sqrt{2} (y - 3) x = - y^{2} + 6 y - 8 \Rightarrow$ $x = \frac{- y^{2} + 6 y - 8}{2 \sqrt{2} (y - 3)} = f^{- 1} (y) \Rightarrow$ $f^{- 1} (x) = \frac{x^{2} - 6 x + 8}{2 \sqrt{2} (3 - x)} .$ ${(f^{- 1} (x))}^{^{'}} = \frac{1}{2 \sqrt{2}} {\frac{(2 x - 6) (3 - x) + (x^{2} - 6 x + 8)}{{(3 - x)}^{2}}}$ $= \frac{1}{2 \sqrt{2}} {\frac{6 x - 2 x^{2} - 18 + 6 x + x^{2} - 6 x + 8}{{(3 - x)}^{2}}}$ $= \frac{1}{2 \sqrt{2}} {\frac{- x^{2} + 6 x - 10}{{(3 - x)}^{2}}}$
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