Q-179570-Posted-by-Infinityaction-30-10-2022-find-the-minimum-of-f-x-f-x-x-2-4-x-2-8x-12-x-25-x-2-4-x-2-16x-16-x-80-f-x-x-4-2-2-x-3-2-

Question Number 179717 by a.lgnaoui last updated on 01/Nov/22

Q 179570 (P o s t e d b y Infinityaction 30.10.2022)

find the minimum of f (x)

f (x) = \sqrt{x^{2} + \frac{4}{x^{2}} - 8 x - \frac{12}{x} + 25} + \sqrt{x^{2} + \frac{4}{x^{2}} - 16 x - \frac{16}{x} + 80}

- - - - - - - - - - - - - - - - - -

f (x) = \sqrt{{(x - 4)}^{2} + {(\frac{2}{x} - 3)}^{2}} + \sqrt{{(x - 8)}^{2} + {(\frac{2}{x} - 4)}^{2}}

D f = R - {0} f (x) ⩾ 0

Min (f (x)) = x / f (x) = 0

\sqrt{{(x - 4)}^{2} + {(\frac{2}{x} - 3)}^{2}} + \sqrt{{(x - 8)}^{2} + {(\frac{2}{x} - 4)}^{2}} = 0

{(x - 4)}^{2} + {(\frac{2}{x} - 3)}^{2} = {(x - 8)}^{2} + {(\frac{2}{x} - 4)}^{2} (1)

x - 8 = (x - 4) - 4 and (\frac{2}{x} - 4) = (\frac{2}{x} - 3) - 1

x^{2} - \frac{55}{8} x + \frac{1}{2} = 0 x = 0, 07351334 a n d x = 6, 80148666}

x = 0, 073513334

f (x) = 49, 044679 (r e j e t e)

x = 6, 80148666 f (x) = 7, 7899055

D o n c 7, 7899055 e s t m i n i m u m d e f (x)

Commented by a.lgnaoui last updated on 01/Nov/22

Q 179570 (l o o k t o a c t u a l a n s e r)

Commented by mr W last updated on 01/Nov/22

w r o n g!

{w h a t}^{'} s t h e r e a s o n t h a t \sqrt{a (x)} + \sqrt{b (x)} = 0

m e a n s m i n i n u m o f \sqrt{a (x)} + \sqrt{b (x)} ?

{i t}^{'} s p a r a d o x t h a t y o u s e t

\sqrt{a (x)} + \sqrt{b (x)} = 0 a n d a t s a m e t i m e y o u

g e t \sqrt{a (x)} + \sqrt{b (x)} = m i n i m u m \neq 0 .

b e s i d e s y o u s e t \sqrt{a (x)} + \sqrt{b (x)} = 0 a n d

a t s a m e t i m e a (x) = b (x) . y o u r l o g i c

i s w e i r d .

Commented by a.lgnaoui last updated on 01/Nov/22

We can find de value by derivation of f (x)

i have trying it but its long

I try aigain I hope that the

resultat were similaire .

Commented by mr W last updated on 01/Nov/22

i t h i n k {i t}^{'} s i m p o s s i b l e t o f i n d t h e

m i n i m u m e x a c t l y i n t h i s q u e s t i o n .

Commented by a.lgnaoui last updated on 01/Nov/22

\sqrt{a {(x)}^{2} + b {(x)}^{2}} + \sqrt{c {(x)}^{2} + d {(x)}^{2}} ⩾ 0 p o u r x \in D f

s o t h e m i n i m u m i s t h e v a l u e o f x

w i c h a {(x)}^{2} + b {(x)}^{2} = 0 a n d c {(x)}^{2} + d {(x)}^{2} = 0

\sqrt{a {(x)}^{2} + b {(x)}^{2}} = - \sqrt{c {(x)}^{2} + d {(x)}^{2}}

a {(x)}^{2} + b {(x)}^{2} = c {(x)}^{2} + d {(x)}^{2}

\lim (f {(x)}_{x \to + \infty} = \lim (f {(x)}_{x \to - \infty}) = \infty

im {(f (x))}_{x \to \pm 0} = + \infty

Commented by mr W last updated on 01/Nov/22

{i t}^{'} s c l e a r l y w r o n g .

j u s t l o o k a t e x a m p l e :

f (x) = \sqrt{x^{2} + 4} + \sqrt{x^{2} + 8}

Answered by manxsol last updated on 01/Nov/22

Commented by manxsol last updated on 01/Nov/22

Commented by mr W last updated on 01/Nov/22

Commented by manxsol last updated on 02/Nov/22

o k, A P \neq P B . u s e f^{'} (x) o \dots \dots .

Commented by manxsol last updated on 02/Nov/22

t h a n k s

Commented by mr W last updated on 02/Nov/22

f^{'} (x) = 0 c a n n o t b e e x a c t l y s o l v e d,

t h e r e f o r e t h e m i n i m u m o f f (x) c a n

n o t b e c a l c u l a t e d e x a c t l y .

Facebook Tweet Pin