f-x-and-g-x-are-functions-with-no-constants-if-f-x-g-x-is-that-mean-f-x-g-x-

Question Number 92379 by M±th+et+s last updated on 06/May/20

f (x) a n d g (x) a r e f u n c t i o n s w i t h n o

c o n s t a n t s .

i f f^{'} (x) = g^{'} (x) i s t h a t m e a n f (x) = g (x)

? ?

Commented by john santu last updated on 06/May/20

no

Commented by mr W last updated on 06/May/20

i f f^{'} (x) = g^{'} (x) t h e n f (x) = g (x) + C

Commented by Prithwish Sen 1 last updated on 06/May/20

I think if there is no constant exists then

if f^{'} (x) = g^{'} (x) \forall x then f (x) = g (x)

but if f^{'} (x) = g^{'} (x) for some finite x then it is

not necessary that f (x) = g (x) . Then it might be

or not might be f (x) = g (x) . Please comment sir .

Commented by Prithwish Sen 1 last updated on 06/May/20

sir what if C = 0

Commented by M±th+et+s last updated on 06/May/20

t h a n k y o u s i r p r i t h w i s h s e n .

t h a t w a s w h a t i m e a n t

Answered by M±th+et+s last updated on 07/May/20

l e t s a y f (x) = {s i n}^{2} (x) a n d g (x) = - {c o s}^{2} (x)

f^{'} (x) = s i n 2 x

b u t f (x) \neq g (x)

s o {i t}^{'} s n o t c o r r e c t

Commented by M±th+et+s last updated on 07/May/20

i d o n t t h i n k t h e r e i s a n o t h e r e x a m p l e

a n d n o w i n o t i c e t h a t i n m y e x a m p l e

f (x) = 1 + g (x) t h a t m e a n s t h e d i f f r e n c e

b e t w e e n t h e m i s a c o n s t a n t

Commented by M±th+et+s last updated on 07/May/20

s o i t h i n k t h a t i f f^{'} (x) = g^{'} (x) t h e n f (x) = g (x)

Commented by Prithwish Sen 1 last updated on 07/May/20

Yes sir . May be (or may be not) it is the only

exception .

Commented by Prithwish Sen 1 last updated on 07/May/20

Excellent! But this two funtions are parallal . Just

a vertical shifting of one function can give you

the other one .

Commented by mr W last updated on 07/May/20

f (x) = {s i n}^{2} (x) a n d g (x) = - {c o s}^{2} (x)

i s j u s t a n e x a m p l e f o r f (x) = g (x) + C

w h e r e C = 1 .

Commented by M±th+et+s last updated on 07/May/20

y e s s i r t h a t i s r i g h t s o w i t h n o c o n s t a n t

(C)

f (x) = g (x)

Commented by mr W last updated on 07/May/20

w h y d o y o u w a n t t o s a y f (x) = g (x)

f r o m f (x) = g (x) + C ?

Commented by M±th+et+s last updated on 07/May/20

i n m y q u e s t i o n i s a i d t h a t n o c o n s t a n t

b e c a u s e i k n o w t h a t f (x) = g (x) + c

b u t i w a s a s k i n g a b o u t a f u n c t i o n s

w i t h n o c o n s t a n t

Commented by mr W last updated on 07/May/20

y o u s h o w e d w i t h e x a m p l e f (x) = \sin^{2} x

a n d g (x) = - \cos^{2} x t h a t t h i s i s n o t

t r u e . t h i s i s c e r t a i n l y n o t t h e o n l y

e x a m p l e .

Commented by M±th+et+s last updated on 07/May/20

y e s b u t l i k e i s a i d i n t h e c o m m e n t s m y

e x a m p l e i s w r o n g b e c a u s e g (x) = - {c o s}^{2} (x)

- {c o s}^{2} (x) = {s i n}^{2} (x) - 1 a n d 1 i s c o n s t a n t