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Question Number 2005 by Rasheed Soomro last updated on 29/Oct/15

If f(x) and g(x) have no constant term then  f ′(x)=g′(x) ⇒ ^(?) f(x)=g(x)?

$${If}\:{f}\left({x}\right)\:{and}\:{g}\left({x}\right)\:{have}\:{no}\:{constant}\:{term}\:{then} \\ $$$${f}\:'\left({x}\right)={g}'\left({x}\right)\overset{?} {\:\Rightarrow\:}{f}\left({x}\right)={g}\left({x}\right)? \\ $$

Commented by prakash jain last updated on 30/Oct/15

If f(x)≠g(x)  since both f(x) and g(x) have no constant  term. f(x)−g(x)=h(x)≠c  f ′(x)−g′(x)=h′(x)≠0  ∴f(x)=g(x)

$$\mathrm{If}\:{f}\left({x}\right)\neq{g}\left({x}\right) \\ $$$$\mathrm{since}\:\mathrm{both}\:{f}\left({x}\right)\:\mathrm{and}\:{g}\left({x}\right)\:\mathrm{have}\:\mathrm{no}\:\mathrm{constant} \\ $$$$\mathrm{term}.\:{f}\left({x}\right)−{g}\left({x}\right)={h}\left({x}\right)\neq{c} \\ $$$${f}\:'\left({x}\right)−{g}'\left({x}\right)={h}'\left({x}\right)\neq\mathrm{0} \\ $$$$\therefore{f}\left({x}\right)={g}\left({x}\right) \\ $$

Commented by Rasheed Soomro last updated on 31/Oct/15

Very Nice!

$$\mathcal{V}{ery}\:\mathcal{N}{ice}! \\ $$

Answered by Filup last updated on 29/Oct/15

No constant, so in form:  f(x)=ax^n   g(x)=bx^m     f′(x)=anx^(n−1)   g′(x)=bmx^(m−1)     f(x)=g(x)  ax^n =bx^m   ∴(ax^n )′=(bx^m )′  anx^(n−1) =bmx^(m−1)     ∴Iff an=bm and n=m will your  statement be correct      (Sorry if this is incorrect)  This is only one solution. I neglected  the use of function such as sine cosine and tangent

$$\mathrm{No}\:\mathrm{constant},\:\mathrm{so}\:\mathrm{in}\:\mathrm{form}: \\ $$$${f}\left({x}\right)={ax}^{{n}} \\ $$$${g}\left({x}\right)={bx}^{{m}} \\ $$$$ \\ $$$${f}'\left({x}\right)={anx}^{{n}−\mathrm{1}} \\ $$$${g}'\left({x}\right)={bmx}^{{m}−\mathrm{1}} \\ $$$$ \\ $$$${f}\left({x}\right)={g}\left({x}\right) \\ $$$${ax}^{{n}} ={bx}^{{m}} \\ $$$$\therefore\left({ax}^{{n}} \right)'=\left({bx}^{{m}} \right)' \\ $$$${anx}^{{n}−\mathrm{1}} ={bmx}^{{m}−\mathrm{1}} \\ $$$$ \\ $$$$\therefore\mathrm{Iff}\:{an}={bm}\:\mathrm{and}\:{n}={m}\:\mathrm{will}\:\mathrm{your} \\ $$$$\mathrm{statement}\:\mathrm{be}\:\mathrm{correct} \\ $$$$ \\ $$$$ \\ $$$$\left({Sorry}\:{if}\:{this}\:{is}\:{incorrect}\right) \\ $$$${This}\:{is}\:{only}\:{one}\:{solution}.\:{I}\:{neglected} \\ $$$${the}\:{use}\:{of}\:{function}\:{such}\:{as}\:{sine}\:{cosine}\:{and}\:{tangent} \\ $$

Commented by Rasheed Soomro last updated on 29/Oct/15

THANK^S   to attempt the question. You have taken  only a special case of polynomial (Monomial).  In case of polynomial f(x) and g(x) may be taken in  general form:  f(x)=Σ_(i=1) ^(n) a_i x^i     and    g(x)=Σ_(i=1) ^(n) b_i x^i    prevented i to be  zero in order to avoid constant term.

$$\mathcal{THANK}^{\mathcal{S}} \:\:{to}\:{attempt}\:{the}\:{question}.\:{You}\:{have}\:{taken} \\ $$$${only}\:{a}\:{special}\:{case}\:{of}\:{polynomial}\:\left({Monomial}\right). \\ $$$${In}\:{case}\:{of}\:{polynomial}\:{f}\left({x}\right)\:{and}\:{g}\left({x}\right)\:{may}\:{be}\:{taken}\:{in} \\ $$$${general}\:{form}: \\ $$$${f}\left({x}\right)=\underset{{i}=\mathrm{1}} {\overset{{n}} {\Sigma}}{a}_{{i}} {x}^{{i}} \:\:\:\:{and}\:\:\:\:{g}\left({x}\right)=\underset{{i}=\mathrm{1}} {\overset{{n}} {\Sigma}}{b}_{{i}} {x}^{{i}} \:\:\:{prevented}\:{i}\:{to}\:{be} \\ $$$${zero}\:{in}\:{order}\:{to}\:{avoid}\:{constant}\:{term}. \\ $$

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