f-x-x-x-x-x-x-f-5-

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

f (x) = \sqrt{x + \sqrt{x + \sqrt{x + \sqrt{x + \sqrt{x + \dots}}}}}

f^{'} (5) =

Commented by bobhans last updated on 14/Jun/20

let y = \sqrt{x + \sqrt{x + \sqrt{x + \sqrt{x + \sqrt{\dots}}}}}

defined for {\begin{cases} x ⩾ 0 \\ y ⩾ 0 \end{cases}

so y^{2} = x + y; y^{2} - y - x = 0

y = \frac{1 + \sqrt{1 + 4 x}}{2} \Rightarrow y^{'} (x) = \frac{\frac{4}{2 \sqrt{1 + 4 x}}}{2} = \frac{1}{\sqrt{1 + 4 x}}

y^{'} (5) = \frac{1}{\sqrt{21}}

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

a b s o l u t e l y c o r r e c t s i r!

e v e n w i t h o u t k n o w l e d g e a b o u t c a l c u l u s

{i t}^{'} s e a s y t o s e e :

y (5) = \sqrt{5 + \sqrt{5 + \sqrt{5 + \sqrt{\dots}}}} > \sqrt{5} > 0

y (6) = \sqrt{6 + \sqrt{6 + \sqrt{6 + \sqrt{\dots}}}} > \sqrt{5 + \sqrt{5 + \sqrt{5 + \sqrt{\dots}}}}

\frac{y (6) - y (5)}{6 - 5} > 0 \Rightarrow y^{'} (5) > 0

Answered by smridha last updated on 13/Jun/20

l e t f (x) = y

s o y^{2} = x + y \dots . (i)

d i f f w r t x w e g e t

2 y \frac{d y}{d x} = 1 + \frac{d y}{d x}, s o y^{^{'}} = \frac{1}{2 y - 1} \dots (i i)

f r o m (i) w e g e t : y = \frac{1 \underline{+} \sqrt{1 + 4 x}}{2}

s o i f x = 5 t h e n : y = \frac{1 \underline{+} \sqrt{21}}{2} .

n o w f r o m (i i) :

y^{^{'}} (5) = \underline{+} \frac{1}{\sqrt{21}}

Commented by bemath last updated on 13/Jun/20

yeahhh \dots

Answered by abdomathmax last updated on 13/Jun/20

f^{2} (x) = x + f (x) \Rightarrow f^{2} (x) - f (x) - x = 0 with f ⩾ 0 and x ⩾ 0

Δ = 1 + 4 x \Rightarrow f (x) = \frac{1 + \sqrt{1 + 4 x}}{2} or f (x) = \frac{1 - \sqrt{1 + 4 x}}{2}

\Rightarrow f (x) = \frac{1 + \sqrt{1 + 4 x}}{2} \Rightarrow f^{^{'}} (x) = \frac{1}{2} \times \frac{4}{2 \sqrt{1 + 4 x}}

= \frac{1}{\sqrt{1 + 4 x}} \Rightarrow f^{^{'}} (5) = \frac{1}{\sqrt{21}}

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

i t h i n k x > 0, y > 0

s o t h e r e i s o n l y o n e p o s s i b i l i t y

f (x) = \frac{1 + \sqrt{1 + 4 x}}{2}

f^{'} (5) = \frac{1}{\sqrt{21}}

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

its clear thst f must be ⩾ 0 from its form f (x) = \sqrt{x + \sqrt{\dots}}

Answered by smridha last updated on 13/Jun/20

l e t f (x) = - y

s o y^{2} = x - y

s o y^{^{'}} = \frac{1}{2 y + 1} n o w y = \frac{- 1 \underline{+} \sqrt{1 + 4 x}}{2}

p u t x = 5 w e g e t y = \frac{- 1 \underline{+} \sqrt{21}}{2}

s o y^{^{'}} (5) = \underline{+} \frac{1}{\sqrt{21}}

s o t h e r e i s t w o p o s s i b l e a n s

w h i c h a r e \underline{+} \frac{1}{\sqrt{21}}

s o y^{^{'}} (5) = \underline{+} \frac{1}{\sqrt{21}} a r e 100 % p e r f e c t .

Answered by smridha last updated on 14/Jun/20

o h h t h e n i f x^{2} = 4 t h e n

x = \sqrt{4} a n d x h a s o n l y o n e v a l u e

t h a t i s + 2 w o w t h a t g r e a t v e r y f u n n y

w h a t i s t h e m e a n i n g o f s q u a r e

r o o t \dots I t h i n k i t i s n e w i n v e n s i o n

t h a t^{2} \sqrt{} m e a n s o n l y + .

x i s u n d e r s q u a r e r o o t s o w e

c a n n o t c o n s i d e r e d o n l y p o s s i t i v e

v a l u e \dots i t i s v e r y c h i l d i s h c o m m e n t

i e v e r r e a d \dots

a t f i r s t k n o w t h e m e a n i n g o f

s q u a r e r o o t \dots

\sqrt{5} > 0 w h e n w e c o n s i d e r a n d

t a k e t h e p o s s i t i v e v a l u e

o f \sqrt{5} n o t t h e n e g e t i v e v a l u e

\sqrt{5} \approx \underline{+} 2.236 s o + 2.236 > 0

- 2.236 < 0 .

s o f (x) = \underline{+} y n o t o n l y + y o r - y

s o t h e g e n r e l s i l u t i o n i s

y^{^{'}} (5) = \underline{+} \frac{1}{\sqrt{21}} a l w a y s b e p e r f e c t .

Commented by bobhans last updated on 14/Jun/20

if x^{2} = 4 \Leftrightarrow (x - 2) (x + 2) = 0

so x = \pm 2, but if x = \sqrt{4} = 2

not x = \sqrt{4} = \pm 2 .

Commented by smridha last updated on 14/Jun/20

w e l l \sqrt{4} = \sqrt{i^{4} 4} = i^{2} . 2 = - 2 i t i s a l s o

c o r r e c t . i^{4} = 1

s o \sqrt{i^{4 n + 4} x} = i^{2 n + 2} \sqrt{x}

w h e r e n = 0, 1, 2, 3 \dots . .

a n d i t g i v e s u s + \sqrt{x} f o r n = o d d

a n d - \sqrt{x} f o r n = e v e n

t h i s i s o u r c o n v e n t i o n t h a t

\sqrt{4} c o s i d e r e d a s + 2 b u t i n g e n e r a l

{t h a t}^{'} s n o t t r u e \sqrt{4} = \underline{+} 2 .

i t j u s t d e p e n d o n p h y s i c a l

s e t u a t i o n a n d w h i c h v a l u e w e

w a n t e d t o t a k e . . b u t t h a t d o e s

n o t m e a n t h e r e i s n o n e g e t i v e

v a l u e .

r e m e m b e r f (x) = y f o r s o l v i n g

t h i s w e s q u a r i n g b o t h s i d e s .

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