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Question Number 114167 by mohammad17 last updated on 17/Sep/20

if f(x)=x^2    ,∀x∈R  then   (1) the function is( one −to−one)    (2) the function is not (one−to−one)    chose (1) or (2)

$${if}\:{f}\left({x}\right)={x}^{\mathrm{2}} \:\:\:,\forall{x}\in{R} \\ $$$${then}\: \\ $$$$\left(\mathrm{1}\right)\:{the}\:{function}\:{is}\left(\:{one}\:−{to}−{one}\right) \\ $$$$ \\ $$$$\left(\mathrm{2}\right)\:{the}\:{function}\:{is}\:{not}\:\left({one}−{to}−{one}\right) \\ $$$$ \\ $$$${chose}\:\left(\mathrm{1}\right)\:{or}\:\left(\mathrm{2}\right) \\ $$

Commented by mohammad17 last updated on 17/Sep/20

help me sir

$${help}\:{me}\:{sir} \\ $$

Commented by bobhans last updated on 17/Sep/20

the function one − to−one if   ∀x_1 ,x_2 ∈D_f  ,where x_1 ≠x_2  then f(x_1 )≠f(x_2 )  in your case f(x)=x^2  , put x_1 =2, x_2 =−2  but f(2)=f(−2), hence f(x)=x^2   not one−to−one

$${the}\:{function}\:{one}\:−\:{to}−{one}\:{if}\: \\ $$$$\forall{x}_{\mathrm{1}} ,{x}_{\mathrm{2}} \in{D}_{{f}} \:,{where}\:{x}_{\mathrm{1}} \neq{x}_{\mathrm{2}} \:{then}\:{f}\left({x}_{\mathrm{1}} \right)\neq{f}\left({x}_{\mathrm{2}} \right) \\ $$$${in}\:{your}\:{case}\:{f}\left({x}\right)={x}^{\mathrm{2}} \:,\:{put}\:{x}_{\mathrm{1}} =\mathrm{2},\:{x}_{\mathrm{2}} =−\mathrm{2} \\ $$$${but}\:{f}\left(\mathrm{2}\right)={f}\left(−\mathrm{2}\right),\:{hence}\:{f}\left({x}\right)={x}^{\mathrm{2}} \\ $$$${not}\:{one}−{to}−{one} \\ $$

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