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Question Number 98398 by bemath last updated on 13/Jun/20

f(x) = (√(x+(√(x+(√(x+(√(x+(√(x+...))))))))))  f ′(5) =

$$\mathrm{f}\left(\mathrm{x}\right)\:=\:\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+...}}}}} \\ $$$$\mathrm{f}\:'\left(\mathrm{5}\right)\:=\: \\ $$

Commented by bobhans last updated on 14/Jun/20

let y =(√(x+(√(x+(√(x+(√(x+(√(...))))))))))  defined for  { ((x≥0)),((y≥0)) :}  so y^2 = x+y ; y^2 −y−x=0  y = ((1 + (√(1+4x)))/2) ⇒y′(x)= ((4/(2(√(1+4x))))/2) = (1/(√(1+4x)))  y′(5)=(1/(√(21))) ■

$$\mathrm{let}\:\mathrm{y}\:=\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{...}}}}} \\ $$$$\mathrm{defined}\:\mathrm{for}\:\begin{cases}{\mathrm{x}\geqslant\mathrm{0}}\\{\mathrm{y}\geqslant\mathrm{0}}\end{cases} \\ $$$$\mathrm{so}\:\mathrm{y}^{\mathrm{2}} =\:\mathrm{x}+\mathrm{y}\:;\:\mathrm{y}^{\mathrm{2}} −\mathrm{y}−\mathrm{x}=\mathrm{0} \\ $$$$\mathrm{y}\:=\:\frac{\mathrm{1}\:+\:\sqrt{\mathrm{1}+\mathrm{4x}}}{\mathrm{2}}\:\Rightarrow\mathrm{y}'\left(\mathrm{x}\right)=\:\frac{\frac{\mathrm{4}}{\mathrm{2}\sqrt{\mathrm{1}+\mathrm{4x}}}}{\mathrm{2}}\:=\:\frac{\mathrm{1}}{\sqrt{\mathrm{1}+\mathrm{4x}}} \\ $$$$\mathrm{y}'\left(\mathrm{5}\right)=\frac{\mathrm{1}}{\sqrt{\mathrm{21}}}\:\blacksquare \\ $$

Commented by mr W last updated on 14/Jun/20

absolutely correct sir!  even without knowledge about calculus  it′s easy to see:  y(5)=(√(5+(√(5+(√(5+(√(...))))))))>(√5)>0  y(6)=(√(6+(√(6+(√(6+(√(...))))))))>(√(5+(√(5+(√(5+(√(...))))))))  ((y(6)−y(5))/(6−5))>0 ⇒y′(5)>0

$${absolutely}\:{correct}\:{sir}! \\ $$$${even}\:{without}\:{knowledge}\:{about}\:{calculus} \\ $$$${it}'{s}\:{easy}\:{to}\:{see}: \\ $$$${y}\left(\mathrm{5}\right)=\sqrt{\mathrm{5}+\sqrt{\mathrm{5}+\sqrt{\mathrm{5}+\sqrt{...}}}}>\sqrt{\mathrm{5}}>\mathrm{0} \\ $$$${y}\left(\mathrm{6}\right)=\sqrt{\mathrm{6}+\sqrt{\mathrm{6}+\sqrt{\mathrm{6}+\sqrt{...}}}}>\sqrt{\mathrm{5}+\sqrt{\mathrm{5}+\sqrt{\mathrm{5}+\sqrt{...}}}} \\ $$$$\frac{{y}\left(\mathrm{6}\right)−{y}\left(\mathrm{5}\right)}{\mathrm{6}−\mathrm{5}}>\mathrm{0}\:\Rightarrow{y}'\left(\mathrm{5}\right)>\mathrm{0} \\ $$

Answered by smridha last updated on 13/Jun/20

let f(x)=y  so y^2 =x+y....(i)  diff wrt x we get  2y(dy/dx)=1+(dy/dx),so y^′ =(1/(2y−1))...(ii)  from (i) we get:y=((1+_− (√(1+4x)))/2)  so if x=5 then:y=((1+_− (√(21)))/2).  now from (ii):  y^′ (5)=+_− (1/(√(21)))

$$\boldsymbol{{let}}\:\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)=\boldsymbol{{y}} \\ $$$$\boldsymbol{{so}}\:\boldsymbol{{y}}^{\mathrm{2}} =\boldsymbol{{x}}+\boldsymbol{{y}}....\left(\boldsymbol{{i}}\right) \\ $$$$\boldsymbol{{diff}}\:\boldsymbol{{wrt}}\:\boldsymbol{{x}}\:\boldsymbol{{we}}\:\boldsymbol{{get}} \\ $$$$\mathrm{2}\boldsymbol{{y}}\frac{\boldsymbol{{dy}}}{\boldsymbol{{dx}}}=\mathrm{1}+\frac{\boldsymbol{{dy}}}{\boldsymbol{{dx}}},\boldsymbol{{so}}\:\boldsymbol{{y}}^{'} =\frac{\mathrm{1}}{\mathrm{2}\boldsymbol{{y}}−\mathrm{1}}...\left(\boldsymbol{{ii}}\right) \\ $$$$\boldsymbol{{from}}\:\left(\boldsymbol{{i}}\right)\:\boldsymbol{{we}}\:\boldsymbol{{get}}:\boldsymbol{{y}}=\frac{\mathrm{1}\underset{−} {+}\sqrt{\mathrm{1}+\mathrm{4}\boldsymbol{{x}}}}{\mathrm{2}} \\ $$$$\boldsymbol{{so}}\:\boldsymbol{{if}}\:\boldsymbol{{x}}=\mathrm{5}\:\boldsymbol{{then}}:\boldsymbol{{y}}=\frac{\mathrm{1}\underset{−} {+}\sqrt{\mathrm{21}}}{\mathrm{2}}. \\ $$$$\boldsymbol{{now}}\:\boldsymbol{{from}}\:\left(\boldsymbol{{ii}}\right): \\ $$$$\boldsymbol{{y}}^{'} \left(\mathrm{5}\right)=\underset{−} {+}\frac{\mathrm{1}}{\sqrt{\mathrm{21}}} \\ $$

Commented by bemath last updated on 13/Jun/20

yeahhh...

$$\mathrm{yeahhh}... \\ $$

Answered by abdomathmax last updated on 13/Jun/20

f^2 (x)=x+f(x) ⇒f^2 (x)−f(x)−x=0  with f≥0 and x≥0  Δ =1+4x ⇒f(x) =((1+(√(1+4x)))/2) or f(x)=((1−(√(1+4x)))/2)  ⇒f(x) =((1+(√(1+4x)))/2) ⇒f^′ (x)=(1/2)×(4/(2(√(1+4x))))  =(1/(√(1+4x))) ⇒f^′ (5) =(1/(√(21)))

$$\mathrm{f}^{\mathrm{2}} \left(\mathrm{x}\right)=\mathrm{x}+\mathrm{f}\left(\mathrm{x}\right)\:\Rightarrow\mathrm{f}^{\mathrm{2}} \left(\mathrm{x}\right)−\mathrm{f}\left(\mathrm{x}\right)−\mathrm{x}=\mathrm{0}\:\:\mathrm{with}\:\mathrm{f}\geqslant\mathrm{0}\:\mathrm{and}\:\mathrm{x}\geqslant\mathrm{0} \\ $$$$\Delta\:=\mathrm{1}+\mathrm{4x}\:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)\:=\frac{\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{4x}}}{\mathrm{2}}\:\mathrm{or}\:\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{1}−\sqrt{\mathrm{1}+\mathrm{4x}}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{f}\left(\mathrm{x}\right)\:=\frac{\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{4x}}}{\mathrm{2}}\:\Rightarrow\mathrm{f}^{'} \left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{4}}{\mathrm{2}\sqrt{\mathrm{1}+\mathrm{4x}}} \\ $$$$=\frac{\mathrm{1}}{\sqrt{\mathrm{1}+\mathrm{4x}}}\:\Rightarrow\mathrm{f}^{'} \left(\mathrm{5}\right)\:=\frac{\mathrm{1}}{\sqrt{\mathrm{21}}} \\ $$

Commented by mr W last updated on 13/Jun/20

i think x>0, y>0  so there is only one possibility  f(x)=((1+(√(1+4x)))/2)  f ′(5)=(1/(√(21)))

$${i}\:{think}\:{x}>\mathrm{0},\:{y}>\mathrm{0} \\ $$$${so}\:{there}\:{is}\:{only}\:{one}\:{possibility} \\ $$$${f}\left({x}\right)=\frac{\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{4}{x}}}{\mathrm{2}} \\ $$$${f}\:'\left(\mathrm{5}\right)=\frac{\mathrm{1}}{\sqrt{\mathrm{21}}} \\ $$

Commented by mathmax by abdo last updated on 13/Jun/20

its clear thst f must be ≥0 from its form f(x)=(√(x+(√(...))))

$$\mathrm{its}\:\mathrm{clear}\:\mathrm{thst}\:\mathrm{f}\:\mathrm{must}\:\mathrm{be}\:\geqslant\mathrm{0}\:\mathrm{from}\:\mathrm{its}\:\mathrm{form}\:\mathrm{f}\left(\mathrm{x}\right)=\sqrt{\mathrm{x}+\sqrt{...}} \\ $$

Answered by smridha last updated on 13/Jun/20

let f(x)=−y  so y^2 =x−y  so y^′ =(1/(2y+1))  now y=((−1+_− (√(1+4x)))/2)  put x=5 we get y=((−1+_− (√(21)))/2)  so y^′ (5)=+_− (1/(√(21)))  so there is two possible ans  which are +_− (1/(√(21)))    so y^′ (5)=+_− (1/(√(21))) are 100% perfect.

$$\boldsymbol{{let}}\:\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)=−\boldsymbol{{y}} \\ $$$$\boldsymbol{{so}}\:\boldsymbol{{y}}^{\mathrm{2}} =\boldsymbol{{x}}−\boldsymbol{{y}} \\ $$$$\boldsymbol{{so}}\:\boldsymbol{{y}}^{'} =\frac{\mathrm{1}}{\mathrm{2}\boldsymbol{{y}}+\mathrm{1}}\:\:\boldsymbol{{now}}\:\boldsymbol{{y}}=\frac{−\mathrm{1}\underset{−} {+}\sqrt{\mathrm{1}+\mathrm{4}\boldsymbol{{x}}}}{\mathrm{2}} \\ $$$$\boldsymbol{{put}}\:\boldsymbol{{x}}=\mathrm{5}\:\boldsymbol{{we}}\:\boldsymbol{{get}}\:\boldsymbol{{y}}=\frac{−\mathrm{1}\underset{−} {+}\sqrt{\mathrm{21}}}{\mathrm{2}} \\ $$$$\boldsymbol{{so}}\:\boldsymbol{{y}}^{'} \left(\mathrm{5}\right)=\underset{−} {+}\frac{\mathrm{1}}{\sqrt{\mathrm{21}}} \\ $$$$\boldsymbol{{so}}\:\boldsymbol{{there}}\:\boldsymbol{{is}}\:\boldsymbol{{two}}\:\boldsymbol{{possible}}\:\boldsymbol{{ans}} \\ $$$$\boldsymbol{{which}}\:\boldsymbol{{are}}\:\underset{−} {+}\frac{\mathrm{1}}{\sqrt{\mathrm{21}}} \\ $$$$\:\:\boldsymbol{{so}}\:\boldsymbol{{y}}^{'} \left(\mathrm{5}\right)=\underset{−} {+}\frac{\mathrm{1}}{\sqrt{\mathrm{21}}}\:\boldsymbol{{are}}\:\mathrm{100\%}\:\boldsymbol{{perfect}}. \\ $$

Answered by smridha last updated on 14/Jun/20

ohh then if x^2 =4 then   x=(√4) and x has only one value  that is+ 2 wow that great very funny  what is the meaning of square  root...I think it is new invension  that^2 (√(  ))   means only +.  x is under square root so we  cannot considered only possitive  value...it is very childish comment  i ever read...  at first know the meaning of  square root...  (√5) >0 when we consider and  take the possitive value  of (√5) not the negetive value  (√5)≈+_− 2.236 so +2.236>0   −2.236<0.  so f(x)=+_− y not only +y or −y  so the genrel silution is  y^′ (5)=+_− (1/(√(21))) always be perfect.

$$\boldsymbol{{ohh}}\:\boldsymbol{{then}}\:\boldsymbol{{if}}\:\boldsymbol{{x}}^{\mathrm{2}} =\mathrm{4}\:\boldsymbol{{then}}\: \\ $$$$\boldsymbol{{x}}=\sqrt{\mathrm{4}}\:\boldsymbol{{and}}\:\boldsymbol{{x}}\:\boldsymbol{{has}}\:\boldsymbol{{only}}\:\boldsymbol{{one}}\:\boldsymbol{{value}} \\ $$$$\boldsymbol{{t}}{ha}\boldsymbol{{t}}\:\boldsymbol{{is}}+\:\mathrm{2}\:\boldsymbol{{wow}}\:\boldsymbol{{that}}\:\boldsymbol{{great}}\:\boldsymbol{{very}}\:\boldsymbol{{funny}} \\ $$$$\boldsymbol{{what}}\:\boldsymbol{{is}}\:\boldsymbol{{the}}\:\boldsymbol{{meaning}}\:\boldsymbol{{of}}\:\boldsymbol{{square}} \\ $$$$\boldsymbol{{root}}...\boldsymbol{{I}}\:\boldsymbol{{think}}\:\boldsymbol{{it}}\:\boldsymbol{{is}}\:\boldsymbol{{new}}\:\boldsymbol{{invension}} \\ $$$$\boldsymbol{{that}}\:^{\mathrm{2}} \sqrt{\:\:}\:\:\:\boldsymbol{{means}}\:\boldsymbol{{only}}\:+. \\ $$$$\boldsymbol{{x}}\:\boldsymbol{{is}}\:\boldsymbol{{under}}\:\boldsymbol{{square}}\:\boldsymbol{{root}}\:\boldsymbol{{so}}\:\boldsymbol{{we}} \\ $$$$\boldsymbol{{cannot}}\:\boldsymbol{{considered}}\:\boldsymbol{{only}}\:\boldsymbol{{possitive}} \\ $$$$\boldsymbol{{value}}...\boldsymbol{{it}}\:\boldsymbol{{is}}\:\boldsymbol{{very}}\:\boldsymbol{{childish}}\:\boldsymbol{{comment}} \\ $$$$\boldsymbol{{i}}\:\boldsymbol{{ever}}\:\boldsymbol{{read}}... \\ $$$$\boldsymbol{{at}}\:\boldsymbol{{first}}\:\boldsymbol{{know}}\:\boldsymbol{{the}}\:\boldsymbol{{meaning}}\:\boldsymbol{{of}} \\ $$$$\boldsymbol{{square}}\:\boldsymbol{{root}}... \\ $$$$\sqrt{\mathrm{5}}\:>\mathrm{0}\:\boldsymbol{{when}}\:\boldsymbol{{we}}\:\boldsymbol{{consider}}\:\boldsymbol{{and}} \\ $$$$\boldsymbol{{take}}\:\boldsymbol{{the}}\:\boldsymbol{{possitive}}\:\boldsymbol{{value}} \\ $$$$\boldsymbol{{of}}\:\sqrt{\mathrm{5}}\:\boldsymbol{{not}}\:\boldsymbol{{the}}\:\boldsymbol{{negetive}}\:\boldsymbol{{value}} \\ $$$$\sqrt{\mathrm{5}}\approx\underset{−} {+}\mathrm{2}.\mathrm{236}\:{so}\:+\mathrm{2}.\mathrm{236}>\mathrm{0}\: \\ $$$$−\mathrm{2}.\mathrm{236}<\mathrm{0}. \\ $$$$\boldsymbol{{so}}\:\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)=\underset{−} {+}\boldsymbol{{y}}\:\boldsymbol{{not}}\:\boldsymbol{{only}}\:+\boldsymbol{{y}}\:\boldsymbol{{or}}\:−\boldsymbol{{y}} \\ $$$$\boldsymbol{{so}}\:\boldsymbol{{the}}\:\boldsymbol{{genrel}}\:\boldsymbol{{silution}}\:\boldsymbol{{is}} \\ $$$$\boldsymbol{{y}}^{'} \left(\mathrm{5}\right)=\underset{−} {+}\frac{\mathrm{1}}{\sqrt{\mathrm{21}}}\:\boldsymbol{{always}}\:\boldsymbol{{be}}\:\boldsymbol{{perfect}}. \\ $$

Commented by bobhans last updated on 14/Jun/20

if x^2 =4 ⇔(x−2)(x+2)=0  so x = ± 2 ,  but  if x = (√4) = 2  not x = (√4) = ± 2 .

$$\mathrm{if}\:\mathrm{x}^{\mathrm{2}} =\mathrm{4}\:\Leftrightarrow\left(\mathrm{x}−\mathrm{2}\right)\left(\mathrm{x}+\mathrm{2}\right)=\mathrm{0} \\ $$$$\mathrm{so}\:\mathrm{x}\:=\:\pm\:\mathrm{2}\:,\:\:\mathrm{but}\:\:\mathrm{if}\:\mathrm{x}\:=\:\sqrt{\mathrm{4}}\:=\:\mathrm{2} \\ $$$$\mathrm{not}\:\mathrm{x}\:=\:\sqrt{\mathrm{4}}\:=\:\pm\:\mathrm{2}\:.\: \\ $$

Commented by smridha last updated on 14/Jun/20

well (√4)=(√(i^4 4))=i^2 .2=−2 it is also  correct .i^4 =1  so (√(i^(4n+4) x))=i^(2n+2) (√x)  where n=0 ,1,2,3.....  and it gives us +(√x) for n=odd  and −(√(x )) for n=even  this is our convention that  (√4) cosidered as +2 but ingeneral  that′s not true (√4)=+_− 2.  it just depend on physical  setuation and which value we  wanted to take..but that does  not mean there is no negetive  value.  remember f(x)=y for solving  this we squaring both sides.

$$\boldsymbol{{well}}\:\sqrt{\mathrm{4}}=\sqrt{\boldsymbol{{i}}^{\mathrm{4}} \mathrm{4}}=\boldsymbol{{i}}^{\mathrm{2}} .\mathrm{2}=−\mathrm{2}\:\boldsymbol{{it}}\:\boldsymbol{{is}}\:\boldsymbol{{also}} \\ $$$$\boldsymbol{{correct}}\:.\boldsymbol{{i}}^{\mathrm{4}} =\mathrm{1} \\ $$$$\boldsymbol{{so}}\:\sqrt{\boldsymbol{{i}}^{\mathrm{4}\boldsymbol{{n}}+\mathrm{4}} \boldsymbol{{x}}}=\boldsymbol{{i}}^{\mathrm{2}\boldsymbol{{n}}+\mathrm{2}} \sqrt{\boldsymbol{{x}}} \\ $$$$\boldsymbol{{where}}\:\boldsymbol{{n}}=\mathrm{0}\:,\mathrm{1},\mathrm{2},\mathrm{3}..... \\ $$$$\boldsymbol{{and}}\:\boldsymbol{{it}}\:\boldsymbol{{gives}}\:\boldsymbol{{us}}\:+\sqrt{\boldsymbol{{x}}}\:\boldsymbol{{for}}\:\boldsymbol{{n}}={odd} \\ $$$$\boldsymbol{{and}}\:−\sqrt{\boldsymbol{{x}}\:}\:\boldsymbol{{for}}\:\boldsymbol{{n}}=\boldsymbol{{even}} \\ $$$$\boldsymbol{{this}}\:\boldsymbol{{is}}\:\boldsymbol{{our}}\:\boldsymbol{{convention}}\:\boldsymbol{{that}} \\ $$$$\sqrt{\mathrm{4}}\:\boldsymbol{{cosidered}}\:\boldsymbol{{as}}\:+\mathrm{2}\:\boldsymbol{{but}}\:\boldsymbol{{ingeneral}} \\ $$$$\boldsymbol{{that}}'\boldsymbol{{s}}\:\boldsymbol{{not}}\:\boldsymbol{{true}}\:\sqrt{\mathrm{4}}=\underset{−} {+}\mathrm{2}. \\ $$$$\boldsymbol{{it}}\:\boldsymbol{{just}}\:\boldsymbol{{depend}}\:\boldsymbol{{on}}\:\boldsymbol{{physical}} \\ $$$$\boldsymbol{{setuation}}\:\boldsymbol{{and}}\:\boldsymbol{{which}}\:\boldsymbol{{value}}\:\boldsymbol{{we}} \\ $$$$\boldsymbol{{wanted}}\:\boldsymbol{{to}}\:\boldsymbol{{take}}..\boldsymbol{{but}}\:\boldsymbol{{that}}\:\boldsymbol{{does}} \\ $$$$\boldsymbol{{not}}\:\boldsymbol{{mean}}\:\boldsymbol{{there}}\:\boldsymbol{{is}}\:\boldsymbol{{no}}\:\boldsymbol{{negetive}} \\ $$$$\boldsymbol{{value}}. \\ $$$$\boldsymbol{{remember}}\:\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)=\boldsymbol{{y}}\:\boldsymbol{{for}}\:\boldsymbol{{solving}} \\ $$$$\boldsymbol{{this}}\:\boldsymbol{{we}}\:\boldsymbol{{squaring}}\:\boldsymbol{{both}}\:\boldsymbol{{sides}}. \\ $$

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