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 ?


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

$i s g = {(1 . 1) . (2.3) . (3 . 5) . (4.7)} a f u n c t i o n ? j u s t i f y i f t h i s i s d e s c r i b e d b y t h e r e l a t i o n g (x) = a x + b t h e n w h a t v a l u e s s h o u l d b e a s s i g n e d t o a a n d b ?$
Commented by mr W last updated on 19/Jan/20

$s i r : p l e a s e {d o n}^{'} t t y p e t h e w h o l e p o s t i n$ $o n e s i n g l e l i n e! {i t}^{'} s t e r r i b l e t o r e a d!$
	Answered by behi83417@gmail.com last updated on 19/Jan/20

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