let f(x)=(√(x+(√(x+1)))) 1) find D_f 2) give the equation of assymtote at point A(0,f(o)) 3) if f(x)∼ a(x−1) +b (x→1) determine a andb 4) calculate f^′ (x) 5) find f^(−1) (x) and (f^(−1) )^′ (x)


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 37892 by abdo mathsup 649 cc last updated on 19/Jun/18

$l e t f (x) = \sqrt{x + \sqrt{x + 1}}$ $1) f i n d D_{f}$ $2) g i v e t h e e q u a t i o n o f a s s y m t o t e a t p o i n t$ $A (0, f (o))$ $3) i f f (x) \sim a (x - 1) + b (x \to 1) d e t e r m i n e a a n d b$ $4) c a l c u l a t e f^{^{'}} (x)$ $5) f i n d f^{- 1} (x) a n d {(f^{- 1})}^{^{'}} (x)$
Commented by prof Abdo imad last updated on 21/Jun/18
$4)we have f^2 (x)=x +(√(x+1)) ⇒ 2f(x)f^′ (x)= 1 +(1/(2(√(x+1)))) ⇒ 2(√(x+(√(x+1)))) f^′ (x)= 1+(1/(2(√(x+1)))) ⇒ f^′ (x)= (1/(2(√(x+(√(x+1)))))){ 1+(1/(2(√(x+1))))} 5) let f(x)=y ⇔ x=f^(−1) (y) ⇒ (√(x+(√(x+1))))=y ⇒ x+(√(x+1))=y^2 ⇒ (√(x+1))=y^2 −x ⇒x+1=(y^2 −x)^2 =x+1 ⇒ x^2 −2y^2 x +y^4 −x−1=0 ⇒x^2 −(2y^2 +1)x +y^4 −1=0 Δ=(2y^2 +1)^2 −4(y^4 −1) =4y^4 +4y^2 +1−4y^4 +4=4y^2 +5 >0 x_1 =((2y^2 +1 +(√(4y^2 +5)))/2) x_2 = ((2y^2 +1 −(√(4y^2 +5)))/2) but we must have x+1≥0 and y^2 −x ≥0 after verification we find that x=((2y^2 +1−(√(4y^2 +5)))/2) ⇒ f^(−1) (x)=((2x^2 +1 −(√(4x^2 +5)))/2) so (f^(−1) )^′ (x)= 2x −(1/2) ((8x)/(2(√(4x^2 +5)))) =2x −((4x)/(√(4x^2 +5))) .$
$4) w e h a v e f^{2} (x) = x + \sqrt{x + 1} \Rightarrow$ $2 f (x) f^{^{'}} (x) = 1 + \frac{1}{2 \sqrt{x + 1}} \Rightarrow$ $2 \sqrt{x + \sqrt{x + 1}} f^{^{'}} (x) = 1 + \frac{1}{2 \sqrt{x + 1}} \Rightarrow$ $f^{^{'}} (x) = \frac{1}{2 \sqrt{x + \sqrt{x + 1}}} {1 + \frac{1}{2 \sqrt{x + 1}}}$ $5) l e t f (x) = y \Leftrightarrow x = f^{- 1} (y) \Rightarrow$ $\sqrt{x + \sqrt{x + 1}} = y \Rightarrow x + \sqrt{x + 1} = y^{2} \Rightarrow$ $\sqrt{x + 1} = y^{2} - x \Rightarrow x + 1 = {(y^{2} - x)}^{2} = x + 1 \Rightarrow$ $x^{2} - 2 y^{2} x + y^{4} - x - 1 = 0 \Rightarrow x^{2} - (2 y^{2} + 1) x + y^{4} - 1 = 0$ $Δ = {(2 y^{2} + 1)}^{2} - 4 (y^{4} - 1)$ $= 4 y^{4} + 4 y^{2} + 1 - 4 y^{4} + 4 = 4 y^{2} + 5 > 0$ $x_{1} = \frac{2 y^{2} + 1 + \sqrt{4 y^{2} + 5}}{2}$ $x_{2} = \frac{2 y^{2} + 1 - \sqrt{4 y^{2} + 5}}{2} b u t w e m u s t h a v e$ $x + 1 ⩾ 0 a n d y^{2} - x ⩾ 0 a f t e r v e r i f i c a t i o n$ $w e f i n d t h a t x = \frac{2 y^{2} + 1 - \sqrt{4 y^{2} + 5}}{2} \Rightarrow$ $f^{- 1} (x) = \frac{2 x^{2} + 1 - \sqrt{4 x^{2} + 5}}{2} s o$ ${(f^{- 1})}^{^{'}} (x) = 2 x - \frac{1}{2} \frac{8 x}{2 \sqrt{4 x^{2} + 5}}$ $= 2 x - \frac{4 x}{\sqrt{4 x^{2} + 5}} .$
Commented by math khazana by abdo last updated on 21/Jun/18

$1) x \in D_{f} \Leftrightarrow x + 1 ⩾ 0 a n d x + \sqrt{x + 1} ⩾ 0 \Rightarrow$ $x ⩾ - 1 a n d x + \sqrt{x + 1} ⩾ 0 s o i f x ⩾ 0 w e g e t$ $x + \sqrt{x + 1} > 0 i f - 1 ⩽ x ⩽ 0$ $\sqrt{x + 1} + x = \sqrt{x + 1} - (- x) a n d$ ${(\sqrt{x + 1})}^{2} - {(- x)}^{2} = x + 1 - x^{2}$ $= - x^{2} + x + 1 \Rightarrow Δ = 1 - 4 (- 1) = 5$ $x_{1} = \frac{- 1 + \sqrt{5}}{2} {a n d x}_{2} = \frac{- 1 - \sqrt{5}}{2}$ $\sqrt{x + 1} + x = \frac{x + 1 - x^{2}}{\sqrt{x + 1} - x} a n d \sqrt{x + 1} - x > 0 s o$ $\sqrt{1 + x} + x ⩾ 0 \Leftrightarrow x \in [\frac{- 1 - \sqrt{5}}{2}, \frac{- 1 + \sqrt{5}}{2}] b u t$ $[- 1, 0] \subset [\frac{- 1 - \sqrt{5}}{2}, \frac{- 1 + \sqrt{5}}{2}] \Rightarrow D_{f} = [- 1, + \infty [$ $2) y = f^{^{'}} (0) x + f (0) b u t f (x) = \sqrt{x + \sqrt{x + 1}} \Rightarrow$ $f (0) = 1 w e h a v e f^{2} (x) = x + \sqrt{x + 1} \Rightarrow$ $2 f (x) f^{^{'}} (x) = 1 + \frac{1}{2 \sqrt{x + 1}} \Rightarrow 2 f (0) f^{^{'}} (0) = \frac{3}{2} \Rightarrow$ $f^{^{'}} (0) = \frac{3}{4} s o t h e e q u a t i o n o f a s s y m p t o t e i s$ $y = \frac{3}{4} x + 1 .$
Commented by math khazana by abdo last updated on 21/Jun/18

$3) w e h a v e a = f^{^{'}} (1) a n d b = f (1)$ $b = \sqrt{1 + \sqrt{2}}$ $2 f (1) f^{^{'}} (1) = 1 + \frac{1}{2 \sqrt{2}} \Rightarrow f^{^{'}} (1) = \frac{1}{2 \sqrt{1 + \sqrt{2}}} (1 + \frac{1}{2 \sqrt{2}}) = a$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum