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Question Number 49983 by Tawa1 last updated on 12/Dec/18
Find the domain of the function and sketch the graph      f(x)  =   { ((2x − 1,           if        x < 0)),((x^2  ,               if      0 ≤ x ≤ 2)),((1 ,                    if       x ≤ 2)) :}
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{domain}\:\mathrm{of}\:\mathrm{the}\:\mathrm{function}\:\mathrm{and}\:\mathrm{sketch}\:\mathrm{the}\:\mathrm{graph} \\ $$$$\:\:\:\:\mathrm{f}\left(\mathrm{x}\right)\:\:=\:\:\begin{cases}{\mathrm{2x}\:−\:\mathrm{1},\:\:\:\:\:\:\:\:\:\:\:\mathrm{if}\:\:\:\:\:\:\:\:\mathrm{x}\:<\:\mathrm{0}}\\{\mathrm{x}^{\mathrm{2}} \:,\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{if}\:\:\:\:\:\:\mathrm{0}\:\leqslant\:\mathrm{x}\:\leqslant\:\mathrm{2}}\\{\mathrm{1}\:,\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{if}\:\:\:\:\:\:\:\mathrm{x}\:\leqslant\:\mathrm{2}}\end{cases} \\ $$
Answered by kaivan.ahmadi last updated on 15/Dec/18
this is not a function since f(0.5)=0.25 for f(x)=x^2   and f(0.5)=1 for f(x)=1.  in genral the domain of each criterion must have no intersection  otherwise f is not function  (0≤x≤2)∩(x≤2)=[0,2] ≠∅
$$\mathrm{this}\:\mathrm{is}\:\mathrm{not}\:\mathrm{a}\:\mathrm{function}\:\mathrm{since}\:\mathrm{f}\left(\mathrm{0}.\mathrm{5}\right)=\mathrm{0}.\mathrm{25}\:\mathrm{for}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{2}} \\ $$$$\mathrm{and}\:\mathrm{f}\left(\mathrm{0}.\mathrm{5}\right)=\mathrm{1}\:\mathrm{for}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{1}. \\ $$$$\mathrm{in}\:\mathrm{genral}\:\mathrm{the}\:\mathrm{domain}\:\mathrm{of}\:\mathrm{each}\:\mathrm{criterion}\:\mathrm{must}\:\mathrm{have}\:\mathrm{no}\:\mathrm{intersection} \\ $$$$\mathrm{otherwise}\:\mathrm{f}\:\mathrm{is}\:\mathrm{not}\:\mathrm{function} \\ $$$$\left(\mathrm{0}\leqslant\mathrm{x}\leqslant\mathrm{2}\right)\cap\left(\mathrm{x}\leqslant\mathrm{2}\right)=\left[\mathrm{0},\mathrm{2}\right]\:\neq\varnothing \\ $$

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