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Question Number 163098 by amin96 last updated on 03/Jan/22

Answered by Rasheed.Sindhi last updated on 03/Jan/22

f(x+y)=f(x)×f(y) , f^( ′) (0)=2 f ′(x)=? f(x+y)=f(x)×f(y) y=0:f(x)=f(x)×f(0)⇒f(0)=1 f ′(x+y)=f ′(x).f(y)+f(x).f ′(y) y=0: f ′(x)= f ′(x).f(0)+f(x).f ′(0) = f ′(x).1+f(x).2 ⇒2f(x)=0⇒f(x)=0⇒f ′(x)=0

$${f}\left({x}+{y}\right)={f}\left({x}\right)×{f}\left({y}\right)\:,\:{f}^{\:'} \left(\mathrm{0}\right)=\mathrm{2} \\ $$$${f}\:'\left({x}\right)=? \\ $$$$\: \\ $$$${f}\left({x}+{y}\right)={f}\left({x}\right)×{f}\left({y}\right) \\ $$$${y}=\mathrm{0}:{f}\left({x}\right)={f}\left({x}\right)×{f}\left(\mathrm{0}\right)\Rightarrow{f}\left(\mathrm{0}\right)=\mathrm{1} \\ $$$${f}\:'\left({x}+{y}\right)={f}\:'\left({x}\right).{f}\left({y}\right)+{f}\left({x}\right).{f}\:'\left({y}\right) \\ $$$${y}=\mathrm{0}: \\ $$$${f}\:'\left({x}\right)=\:{f}\:'\left({x}\right).{f}\left(\mathrm{0}\right)+{f}\left({x}\right).{f}\:'\left(\mathrm{0}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:=\:{f}\:'\left({x}\right).\mathrm{1}+{f}\left({x}\right).\mathrm{2} \\ $$$$\:\:\:\:\:\Rightarrow\mathrm{2}{f}\left({x}\right)=\mathrm{0}\Rightarrow{f}\left({x}\right)=\mathrm{0}\Rightarrow{f}\:'\left({x}\right)=\mathrm{0} \\ $$

Commented by mr W last updated on 05/Jan/22

i don′t think it′s correct sir. y is not a function of x. both x and y stand just for values here. so f ′(x+y)=f ′(x).f(y)+f(x).f ′(y) is not a valid operation here, i think. besides, with f(x)=f′(x)=0, the conditions f′(0)=2 and f(0)=1 are not fulfilled.

$${i}\:{don}'{t}\:{think}\:{it}'{s}\:{correct}\:{sir}. \\ $$$${y}\:{is}\:{not}\:{a}\:{function}\:{of}\:{x}.\:{both}\:{x}\:{and} \\ $$$${y}\:{stand}\:{just}\:{for}\:{values}\:{here}.\:{so} \\ $$$${f}\:'\left({x}+{y}\right)={f}\:'\left({x}\right).{f}\left({y}\right)+{f}\left({x}\right).{f}\:'\left({y}\right) \\ $$$${is}\:{not}\:{a}\:{valid}\:{operation}\:{here},\:{i}\:{think}. \\ $$$$ \\ $$$${besides},\:{with}\:{f}\left({x}\right)={f}'\left({x}\right)=\mathrm{0},\:{the}\: \\ $$$${conditions}\:{f}'\left(\mathrm{0}\right)=\mathrm{2}\:{and}\:{f}\left(\mathrm{0}\right)=\mathrm{1}\:{are}\:{not} \\ $$$${fulfilled}. \\ $$

Commented by Rasheed.Sindhi last updated on 05/Jan/22

THANX Sir! When you commenting on my post in this way I feel I′m getting ′table tution′ from an expert teacher!

$$\mathcal{THANX}\:\:\mathcal{S}{ir}! \\ $$$${When}\:{you}\:{commenting}\:{on}\:{my}\:{post} \\ $$$${in}\:{this}\:{way}\:{I}\:{feel}\:{I}'{m}\:{getting}\:'{table} \\ $$$${tution}'\:{from}\:{an}\:{expert}\:{teacher}! \\ $$

Commented by mr W last updated on 05/Jan/22

thanks also back to you sir! we have common interest for some topics, so our ways cross somewhere. i′m glad that we can learn from each other.

$${thanks}\:{also}\:{back}\:{to}\:{you}\:{sir}! \\ $$$${we}\:{have}\:{common}\:{interest}\:{for}\:{some} \\ $$$${topics},\:{so}\:{our}\:{ways}\:{cross}\:{somewhere}.\: \\ $$$${i}'{m}\:{glad}\:{that}\:{we}\:{can}\:{learn}\:{from}\: \\ $$$${each}\:{other}. \\ $$

Commented by Rasheed.Sindhi last updated on 05/Jan/22

Like water flows from high position to low position, knowledge transfers from mr W to me!

$${Like}\:{water}\:{flows}\:{from}\:{high}\:{position} \\ $$$${to}\:{low}\:{position},\:{knowledge}\:{transfers} \\ $$$${from}\:{mr}\:{W}\:{to}\:{me}! \\ $$

Answered by mr W last updated on 05/Jan/22

f(x+y)=f(x)×f(y) this is the feature of an exponential function, since a^(m+n) =a^m ×a^n . so assume f(x)=a^x which fulfills f(x+y)=f(x)×f(y). f′(x)=(ln a)a^x f′(0)=(ln a)a^0 =ln a=^! 2 ⇒a=e^2 ⇒f(x)=e^(2x) ⇒f′(x)=2e^(2x)

$${f}\left({x}+{y}\right)={f}\left({x}\right)×{f}\left({y}\right) \\ $$$${this}\:{is}\:{the}\:{feature}\:{of}\:{an}\:{exponential} \\ $$$${function},\:{since}\:{a}^{{m}+{n}} ={a}^{{m}} ×{a}^{{n}} . \\ $$$${so}\:{assume}\:{f}\left({x}\right)={a}^{{x}} \:{which}\:{fulfills} \\ $$$${f}\left({x}+{y}\right)={f}\left({x}\right)×{f}\left({y}\right). \\ $$$${f}'\left({x}\right)=\left(\mathrm{ln}\:{a}\right){a}^{{x}} \\ $$$${f}'\left(\mathrm{0}\right)=\left(\mathrm{ln}\:{a}\right){a}^{\mathrm{0}} =\mathrm{ln}\:{a}\overset{!} {=}\mathrm{2} \\ $$$$\Rightarrow{a}={e}^{\mathrm{2}} \\ $$$$\Rightarrow{f}\left({x}\right)={e}^{\mathrm{2}{x}} \\ $$$$\Rightarrow{f}'\left({x}\right)=\mathrm{2}{e}^{\mathrm{2}{x}} \\ $$

Commented by Rasheed.Sindhi last updated on 05/Jan/22

∨ ∩i⊂∈! ThanX sir!

$$\vee\:\cap\mathrm{i}\subset\in! \\ $$$$\mathcal{T}{han}\mathcal{X}\:\boldsymbol{{sir}}! \\ $$

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