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Question-50010




Question Number 50010 by Tawa1 last updated on 13/Dec/18
Commented by MJS last updated on 13/Dec/18
is it really x≤2 in the 3^(rd)  line?  in this case it′s not defined for x>2 and it′s  f(x)= { ((−∞<x<0: (2x−1)∨1)),((       0≤x≤2: x^2 ∨1)) :}
$$\mathrm{is}\:\mathrm{it}\:\mathrm{really}\:{x}\leqslant\mathrm{2}\:\mathrm{in}\:\mathrm{the}\:\mathrm{3}^{\mathrm{rd}} \:\mathrm{line}? \\ $$$$\mathrm{in}\:\mathrm{this}\:\mathrm{case}\:\mathrm{it}'\mathrm{s}\:\mathrm{not}\:\mathrm{defined}\:\mathrm{for}\:{x}>\mathrm{2}\:\mathrm{and}\:\mathrm{it}'\mathrm{s} \\ $$$${f}\left({x}\right)=\begin{cases}{−\infty<{x}<\mathrm{0}:\:\left(\mathrm{2}{x}−\mathrm{1}\right)\vee\mathrm{1}}\\{\:\:\:\:\:\:\:\mathrm{0}\leqslant{x}\leqslant\mathrm{2}:\:{x}^{\mathrm{2}} \vee\mathrm{1}}\end{cases} \\ $$
Commented by Tawa1 last updated on 13/Dec/18
God bless you sir.  But please help me solve it with   x < 2    if possible
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\:\mathrm{But}\:\mathrm{please}\:\mathrm{help}\:\mathrm{me}\:\mathrm{solve}\:\mathrm{it}\:\mathrm{with}\:\:\:\mathrm{x}\:<\:\mathrm{2}\:\:\:\:\mathrm{if}\:\mathrm{possible} \\ $$

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