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Question Number 84121 by Rio Michael last updated on 09/Mar/20
given  that   g(x) =  { ((x + 2 , if  0 ≤ x < 2)),((x^2  , if  2 ≤ x < 4)) :}  is periodic of period 4.   sketch the curve for g(x) in the interval    0≤ x < 8  evaluate  g(−6).
$$\mathrm{given}\:\:\mathrm{that} \\ $$$$\:\mathrm{g}\left({x}\right)\:=\:\begin{cases}{{x}\:+\:\mathrm{2}\:,\:\mathrm{if}\:\:\mathrm{0}\:\leqslant\:{x}\:<\:\mathrm{2}}\\{{x}^{\mathrm{2}} \:,\:\mathrm{if}\:\:\mathrm{2}\:\leqslant\:{x}\:<\:\mathrm{4}}\end{cases} \\ $$$$\mathrm{is}\:\mathrm{periodic}\:\mathrm{of}\:\mathrm{period}\:\mathrm{4}.\: \\ $$$$\mathrm{sketch}\:\mathrm{the}\:\mathrm{curve}\:\mathrm{for}\:\mathrm{g}\left({x}\right)\:\mathrm{in}\:\mathrm{the}\:\mathrm{interval} \\ $$$$\:\:\mathrm{0}\leqslant\:{x}\:<\:\mathrm{8} \\ $$$$\mathrm{evaluate}\:\:\mathrm{g}\left(−\mathrm{6}\right). \\ $$
Commented by MJS last updated on 09/Mar/20
period 4 ⇒ g(x)=g(x+4z); z∈Z  ⇒ g(−6)=g(−6+4×2)=g(2)=2^2 =4
$$\mathrm{period}\:\mathrm{4}\:\Rightarrow\:{g}\left({x}\right)={g}\left({x}+\mathrm{4}{z}\right);\:{z}\in\mathbb{Z} \\ $$$$\Rightarrow\:{g}\left(−\mathrm{6}\right)={g}\left(−\mathrm{6}+\mathrm{4}×\mathrm{2}\right)={g}\left(\mathrm{2}\right)=\mathrm{2}^{\mathrm{2}} =\mathrm{4} \\ $$
Commented by Rio Michael last updated on 09/Mar/20
thanks sir how about the graph
$$\mathrm{thanks}\:\mathrm{sir}\:\mathrm{how}\:\mathrm{about}\:\mathrm{the}\:\mathrm{graph} \\ $$
Commented by MJS last updated on 09/Mar/20
the graph in [0; 4[ is the same as in [4; 8[  x     g(x)  0     2  .5    2.5  1      3  1.5  3.5  2      4  2.5  6.25  3      9  3.5  12.25  4      16 is not part of g(x) but the lim_(x→4^− )  g(x)            ⇒ the graph seems to reach it but it doesn′t                  mark the point (4∣16) with a “○”  4     2  4.5  2.5  ...  8      16 same as (4∣16)
$$\mathrm{the}\:\mathrm{graph}\:\mathrm{in}\:\left[\mathrm{0};\:\mathrm{4}\left[\:\mathrm{is}\:\mathrm{the}\:\mathrm{same}\:\mathrm{as}\:\mathrm{in}\:\left[\mathrm{4};\:\mathrm{8}\left[\right.\right.\right.\right. \\ $$$${x}\:\:\:\:\:{g}\left({x}\right) \\ $$$$\mathrm{0}\:\:\:\:\:\mathrm{2} \\ $$$$.\mathrm{5}\:\:\:\:\mathrm{2}.\mathrm{5} \\ $$$$\mathrm{1}\:\:\:\:\:\:\mathrm{3} \\ $$$$\mathrm{1}.\mathrm{5}\:\:\mathrm{3}.\mathrm{5} \\ $$$$\mathrm{2}\:\:\:\:\:\:\mathrm{4} \\ $$$$\mathrm{2}.\mathrm{5}\:\:\mathrm{6}.\mathrm{25} \\ $$$$\mathrm{3}\:\:\:\:\:\:\mathrm{9} \\ $$$$\mathrm{3}.\mathrm{5}\:\:\mathrm{12}.\mathrm{25} \\ $$$$\mathrm{4}\:\:\:\:\:\:\mathrm{16}\:\mathrm{is}\:\mathrm{not}\:\mathrm{part}\:\mathrm{of}\:{g}\left({x}\right)\:\mathrm{but}\:\mathrm{the}\:\underset{{x}\rightarrow\mathrm{4}^{−} } {\mathrm{lim}}\:{g}\left({x}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\mathrm{the}\:\mathrm{graph}\:\mathrm{seems}\:\mathrm{to}\:\mathrm{reach}\:\mathrm{it}\:\mathrm{but}\:\mathrm{it}\:\mathrm{doesn}'\mathrm{t} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{mark}\:\mathrm{the}\:\mathrm{point}\:\left(\mathrm{4}\mid\mathrm{16}\right)\:\mathrm{with}\:\mathrm{a}\:“\circ'' \\ $$$$\mathrm{4}\:\:\:\:\:\mathrm{2} \\ $$$$\mathrm{4}.\mathrm{5}\:\:\mathrm{2}.\mathrm{5} \\ $$$$… \\ $$$$\mathrm{8}\:\:\:\:\:\:\mathrm{16}\:\mathrm{same}\:\mathrm{as}\:\left(\mathrm{4}\mid\mathrm{16}\right) \\ $$
Commented by Rio Michael last updated on 10/Mar/20
thanks sir
$$\mathrm{thanks}\:\mathrm{sir} \\ $$

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