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

f(x) and g(x) are functions with no constants. if f ′(x)=g ′(x) is that mean f(x)=g(x) ??

$${f}\left({x}\right)\:{and}\:{g}\left({x}\right)\:{are}\:{functions}\:{with}\:{no} \\ $$$${constants}. \\ $$$${if}\:{f}\:'\left({x}\right)={g}\:'\left({x}\right)\:{is}\:{that}\:{mean}\:{f}\left({x}\right)={g}\left({x}\right) \\ $$$$?? \\ $$

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

no

$$\mathrm{no} \\ $$

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

if f ′(x)=g ′(x) then f(x)=g(x)+C

$${if}\:{f}\:'\left({x}\right)={g}\:'\left({x}\right)\:{then}\:{f}\left({x}\right)={g}\left({x}\right)+{C} \\ $$

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) ∀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.

$$\mathrm{I}\:\mathrm{think}\:\mathrm{if}\:\mathrm{there}\:\mathrm{is}\:\mathrm{no}\:\mathrm{constant}\:\mathrm{exists}\:\mathrm{then}\: \\ $$$$\mathrm{if}\:\mathrm{f}'\left(\mathrm{x}\right)=\mathrm{g}'\left(\mathrm{x}\right)\:\forall\mathrm{x}\:\mathrm{then}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{g}\left(\mathrm{x}\right) \\ $$$$\mathrm{but}\:\mathrm{if}\:\mathrm{f}'\left(\mathrm{x}\right)\:=\:\mathrm{g}'\left(\mathrm{x}\right)\:\mathrm{for}\:\mathrm{some}\:\mathrm{finite}\:\mathrm{x}\:\mathrm{then}\:\mathrm{it}\:\mathrm{is} \\ $$$$\mathrm{not}\:\mathrm{necessary}\:\mathrm{that}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{g}\left(\mathrm{x}\right)\:.\:\mathrm{Then}\:\mathrm{it}\:\mathrm{might}\:\mathrm{be} \\ $$$$\mathrm{or}\:\mathrm{not}\:\mathrm{might}\:\mathrm{be}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{g}\left(\mathrm{x}\right).\:\mathrm{Please}\:\mathrm{comment}\:\mathrm{sir}. \\ $$

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

sir what if C = 0

$$\mathrm{sir}\:\mathrm{what}\:\mathrm{if}\:\mathrm{C}\:=\:\mathrm{0} \\ $$

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

thank you sir prithwish sen. that was what i meant

$${thank}\:{you}\:{sir}\:{prithwish}\:{sen}. \\ $$$${that}\:{was}\:{what}\:{i}\:{meant} \\ $$

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

let say f(x)=sin^2 (x) and g(x)=−cos^2 (x) f′(x)=sin2x but f(x)≠g(x) so it′s not correct

$${let}\:{say}\:{f}\left({x}\right)={sin}^{\mathrm{2}} \left({x}\right)\:{and}\:{g}\left({x}\right)=−{cos}^{\mathrm{2}} \left({x}\right) \\ $$$${f}'\left({x}\right)={sin}\mathrm{2}{x}\: \\ $$$${but}\:{f}\left({x}\right)\neq{g}\left({x}\right) \\ $$$${so}\:{it}'{s}\:{not}\:{correct} \\ $$

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

i dont think there is another example and now inotice that in my example f(x)=1+g(x) that means the diffrence between them is a constant

$${i}\:{dont}\:{think}\:{there}\:{is}\:{another}\:{example} \\ $$$${and}\:{now}\:{inotice}\:{that}\:{in}\:{my}\:{example} \\ $$$${f}\left({x}\right)=\mathrm{1}+{g}\left({x}\right)\:{that}\:{means}\:{the}\:{diffrence} \\ $$$${between}\:{them}\:{is}\:{a}\:{constant} \\ $$

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

so i think that if f ′(x)=g ′(x) then f(x)=g(x)

$${so}\:{i}\:{think}\:{that}\:{if}\:{f}\:'\left({x}\right)={g}\:'\left({x}\right)\:{then}\:{f}\left({x}\right)={g}\left({x}\right) \\ $$

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

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

$$\mathrm{Yes}\:\mathrm{sir}.\:\mathrm{May}\:\mathrm{be}\:\left(\mathrm{or}\:\mathrm{may}\:\mathrm{be}\:\mathrm{not}\right)\:\mathrm{it}\:\mathrm{is}\:\mathrm{the}\:\mathrm{only}\: \\ $$$$\mathrm{exception}. \\ $$

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.

$$\mathrm{Excellent}\:!\:\mathrm{But}\:\mathrm{this}\:\mathrm{two}\:\mathrm{funtions}\:\mathrm{are}\:\mathrm{parallal}.\:\mathrm{Just} \\ $$$$\mathrm{a}\:\mathrm{vertical}\:\mathrm{shifting}\:\mathrm{of}\:\mathrm{one}\:\mathrm{function}\:\mathrm{can}\:\mathrm{give}\:\mathrm{you} \\ $$$$\mathrm{the}\:\mathrm{other}\:\mathrm{one}.\: \\ $$

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

f(x)=sin^2 (x) and g(x)=−cos^2 (x) is just an example for f(x)=g(x)+C where C=1.

$${f}\left({x}\right)={sin}^{\mathrm{2}} \left({x}\right)\:{and}\:{g}\left({x}\right)=−{cos}^{\mathrm{2}} \left({x}\right) \\ $$$${is}\:{just}\:{an}\:{example}\:{for}\:{f}\left({x}\right)={g}\left({x}\right)+{C} \\ $$$${where}\:{C}=\mathrm{1}. \\ $$

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

yes sir that is right so with no constant (C) f(x)=g(x)

$${yes}\:{sir}\:{that}\:{is}\:{right}\:{so}\:{with}\:{no}\:{constant} \\ $$$$\left({C}\right)\:\:\:\: \\ $$$${f}\left({x}\right)={g}\left({x}\right) \\ $$

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

why do you want to say f(x)=g(x) from f(x)=g(x)+C?

$${why}\:{do}\:{you}\:{want}\:{to}\:{say}\:{f}\left({x}\right)={g}\left({x}\right) \\ $$$${from}\:{f}\left({x}\right)={g}\left({x}\right)+{C}? \\ $$

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

in my question i said that no constant because i know that f(x)=g(x)+c but i was asking about a functions with no constant

$${in}\:{my}\:{question}\:{i}\:{said}\:{that}\:{no}\:{constant} \\ $$$${because}\:{i}\:{know}\:{that}\:{f}\left({x}\right)={g}\left({x}\right)+{c} \\ $$$${but}\:{i}\:{was}\:{asking}\:{about}\:{a}\:{functions}\: \\ $$$${with}\:{no}\:{constant}\: \\ $$

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

you showed with example f(x)=sin^2 x and g(x)=−cos^2 x that this is not true. this is certainly not the only example.

$${you}\:{showed}\:{with}\:{example}\:{f}\left({x}\right)=\mathrm{sin}^{\mathrm{2}} \:{x} \\ $$$${and}\:{g}\left({x}\right)=−\mathrm{cos}^{\mathrm{2}} \:{x}\:{that}\:{this}\:{is}\:{not} \\ $$$${true}.\:{this}\:{is}\:{certainly}\:{not}\:{the}\:{only} \\ $$$${example}. \\ $$

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

yes but like i said in the comments my example is wrong because g(x)=−cos^2 (x) −cos^2 (x)= sin^2 (x)−1 and 1 is constant

$${yes}\:{but}\:{like}\:{i}\:{said}\:{in}\:{the}\:{comments}\:{my} \\ $$$${example}\:{is}\:{wrong}\:{because}\:{g}\left({x}\right)=−{cos}^{\mathrm{2}} \left({x}\right) \\ $$$$−{cos}^{\mathrm{2}} \left({x}\right)=\:{sin}^{\mathrm{2}} \left({x}\right)−\mathrm{1}\:\:{and}\:\mathrm{1}\:{is}\:{constant} \\ $$

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