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Question Number 78685 by berket last updated on 19/Jan/20

is g ={(1. 1).(2.3).(3 .5).(4.7)} a function? justify if this is described by the relation g(x)=ax+b then what values should be assigned to a and b ?

$${is}\:{g}\:=\left\{\left(\mathrm{1}.\:\mathrm{1}\right).\left(\mathrm{2}.\mathrm{3}\right).\left(\mathrm{3}\:.\mathrm{5}\right).\left(\mathrm{4}.\mathrm{7}\right)\right\}\:{a}\:{function}?\:{justify}\:{if}\:{this}\:{is}\:{described}\:{by}\:{the}\:{relation}\:{g}\left({x}\right)={ax}+{b}\:{then}\:{what}\:{values}\:{should}\:{be}\:{assigned}\:{to}\:{a}\:{and}\:{b}\:? \\ $$

Commented by mr W last updated on 19/Jan/20

sir: please don′t type the whole post in one single line! it′s terrible to read!

$${sir}:\:{please}\:{don}'{t}\:{type}\:{the}\:{whole}\:{post}\:{in} \\ $$$${one}\:{single}\:{line}!\:{it}'{s}\:{terrible}\:{to}\:{read}! \\ $$

Answered by behi83417@gmail.com last updated on 19/Jan/20

yes. it is a function. notice the: x′s in the :(x,y). 1.if the x′s are different,then the f,is function. 2.if two x′s are equail,notice to:y′s. if y′s are equail too,then the f,is function. if y′s not equail,then the f,is not a function. let:g(x)=ax+b g(3)=5⇒5=a×3+b=3a+b g(2)=3⇒3=a×2+b=2a+b ⇒ { ((3a+b=5)),((2a+b=3)) :}⇒a=2,b=−1⇒g(x)=2x−1 but: (1,1) not works with this relation. I think it is :(1,0) inested of: (1,1)

$$\mathrm{yes}.\:\mathrm{it}\:\mathrm{is}\:\mathrm{a}\:\mathrm{function}. \\ $$$$\mathrm{notice}\:\mathrm{the}:\:\boldsymbol{\mathrm{x}}'\mathrm{s}\:\mathrm{in}\:\mathrm{the}\::\left(\mathrm{x},\mathrm{y}\right). \\ $$$$\mathrm{1}.\mathrm{if}\:\mathrm{the}\:\boldsymbol{\mathrm{x}}'\mathrm{s}\:\mathrm{are}\:\mathrm{different},\mathrm{then}\:\mathrm{the}\:\boldsymbol{\mathrm{f}},\mathrm{is}\: \\ $$$$\mathrm{function}. \\ $$$$\mathrm{2}.\mathrm{if}\:\mathrm{two}\:\boldsymbol{\mathrm{x}}'\mathrm{s}\:\mathrm{are}\:\mathrm{equail},\mathrm{notice}\:\mathrm{to}:\boldsymbol{\mathrm{y}}'\mathrm{s}. \\ $$$$\mathrm{if}\:\boldsymbol{\mathrm{y}}'\mathrm{s}\:\mathrm{are}\:\mathrm{equail}\:\mathrm{too},\mathrm{then}\:\mathrm{the}\:\boldsymbol{\mathrm{f}},\mathrm{is}\:\mathrm{function}. \\ $$$$\mathrm{if}\:\boldsymbol{\mathrm{y}}'\mathrm{s}\:\mathrm{not}\:\mathrm{equail},\mathrm{then}\:\mathrm{the}\:\boldsymbol{\mathrm{f}},\mathrm{is}\:\mathrm{not}\:\mathrm{a}\:\mathrm{function}. \\ $$$$\mathrm{let}:\mathrm{g}\left(\mathrm{x}\right)=\mathrm{ax}+\mathrm{b} \\ $$$$\mathrm{g}\left(\mathrm{3}\right)=\mathrm{5}\Rightarrow\mathrm{5}=\mathrm{a}×\mathrm{3}+\mathrm{b}=\mathrm{3a}+\mathrm{b} \\ $$$$\mathrm{g}\left(\mathrm{2}\right)=\mathrm{3}\Rightarrow\mathrm{3}=\mathrm{a}×\mathrm{2}+\mathrm{b}=\mathrm{2a}+\mathrm{b} \\ $$$$\Rightarrow\begin{cases}{\mathrm{3a}+\mathrm{b}=\mathrm{5}}\\{\mathrm{2a}+\mathrm{b}=\mathrm{3}}\end{cases}\Rightarrow\mathrm{a}=\mathrm{2},\mathrm{b}=−\mathrm{1}\Rightarrow\mathrm{g}\left(\mathrm{x}\right)=\mathrm{2x}−\mathrm{1} \\ $$$$\mathrm{but}:\:\:\left(\mathrm{1},\mathrm{1}\right)\:\mathrm{not}\:\mathrm{works}\:\mathrm{with}\:\mathrm{this}\:\mathrm{relation}. \\ $$$$\mathrm{I}\:\mathrm{think}\:\mathrm{it}\:\mathrm{is}\::\left(\mathrm{1},\mathrm{0}\right)\:\mathrm{inested}\:\mathrm{of}:\:\left(\mathrm{1},\mathrm{1}\right) \\ $$

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