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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 byMJS 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 byRio 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 byMJS 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 ♮○ε 4 2 4.5 2.5 ... 8 16 same as (4∣16)

$$\mathrm{the}\:\mathrm{graph}\:\mathrm{in}\:\left[\mathrm{0};\:\mathrm{4}\left[\:\mathrm{is}\:\mathrm{the}\:\mathrm{same}\:\mathrm{as}\:\mathrm{in}\:\left[\mathrm{4};\:\mathrm{8}\left[\right.\right.\right.\right. \\ $$ $${x}\:\:\:\:\:{g}\left({x}\right) \\ $$ $$\mathrm{0}\:\:\:\:\:\mathrm{2} \\ $$ $$.\mathrm{5}\:\:\:\:\mathrm{2}.\mathrm{5} \\ $$ $$\mathrm{1}\:\:\:\:\:\:\mathrm{3} \\ $$ $$\mathrm{1}.\mathrm{5}\:\:\mathrm{3}.\mathrm{5} \\ $$ $$\mathrm{2}\:\:\:\:\:\:\mathrm{4} \\ $$ $$\mathrm{2}.\mathrm{5}\:\:\mathrm{6}.\mathrm{25} \\ $$ $$\mathrm{3}\:\:\:\:\:\:\mathrm{9} \\ $$ $$\mathrm{3}.\mathrm{5}\:\:\mathrm{12}.\mathrm{25} \\ $$ $$\mathrm{4}\:\:\:\:\:\:\mathrm{16}\:\mathrm{is}\:\mathrm{not}\:\mathrm{part}\:\mathrm{of}\:{g}\left({x}\right)\:\mathrm{but}\:\mathrm{the}\:\underset{{x}\rightarrow\mathrm{4}^{−} } {\mathrm{lim}}\:{g}\left({x}\right) \\ $$ $$\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\mathrm{the}\:\mathrm{graph}\:\mathrm{seems}\:\mathrm{to}\:\mathrm{reach}\:\mathrm{it}\:\mathrm{but}\:\mathrm{it}\:\mathrm{doesn}'\mathrm{t} \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{mark}\:\mathrm{the}\:\mathrm{point}\:\left(\mathrm{4}\mid\mathrm{16}\right)\:\mathrm{with}\:\mathrm{a}\:\natural\circ\varepsilon \\ $$ $$\mathrm{4}\:\:\:\:\:\mathrm{2} \\ $$ $$\mathrm{4}.\mathrm{5}\:\:\mathrm{2}.\mathrm{5} \\ $$ $$... \\ $$ $$\mathrm{8}\:\:\:\:\:\:\mathrm{16}\:\mathrm{same}\:\mathrm{as}\:\left(\mathrm{4}\mid\mathrm{16}\right) \\ $$

Commented byRio Michael last updated on 10/Mar/20

thanks sir

$$\mathrm{thanks}\:\mathrm{sir} \\ $$

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