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Question Number 67684 by Rio Michael last updated on 30/Aug/19

given the function   f(x) = { ((x^2   , for   0≤ x< 2)),((ax + 3, for  2≤ x < 4)) :}  is periodic of period  4, and is continuous.  a) Find  the value of  a.  b) Find the valu of  f(6)  c) sketch the graph for y =f(x).  help me please, for the graph i don′t know wbere to put  y=x^2  and y = ax + 3 and  where do i put a closed  dot and an open dot.

$${given}\:{the}\:{function}\: \\ $$ $${f}\left({x}\right)\:=\begin{cases}{{x}^{\mathrm{2}} \:\:,\:{for}\:\:\:\mathrm{0}\leqslant\:{x}<\:\mathrm{2}}\\{{ax}\:+\:\mathrm{3},\:{for}\:\:\mathrm{2}\leqslant\:{x}\:<\:\mathrm{4}}\end{cases} \\ $$ $${is}\:{periodic}\:{of}\:{period}\:\:\mathrm{4},\:{and}\:{is}\:{continuous}. \\ $$ $$\left.{a}\right)\:{Find}\:\:{the}\:{value}\:{of}\:\:{a}. \\ $$ $$\left.{b}\right)\:{Find}\:{the}\:{valu}\:{of}\:\:{f}\left(\mathrm{6}\right) \\ $$ $$\left.{c}\right)\:{sketch}\:{the}\:{graph}\:{for}\:{y}\:={f}\left({x}\right). \\ $$ $${help}\:{me}\:{please},\:{for}\:{the}\:{graph}\:{i}\:{don}'{t}\:{know}\:{wbere}\:{to}\:{put}\:\:{y}={x}^{\mathrm{2}} \:{and}\:{y}\:=\:{ax}\:+\:\mathrm{3}\:{and} \\ $$ $${where}\:{do}\:{i}\:{put}\:{a}\:{closed}\:\:{dot}\:{and}\:{an}\:{open}\:{dot}. \\ $$ $$ \\ $$

Commented byMJS last updated on 30/Aug/19

if f(x) is continuous within [0; 4] ⇒  ⇒ x^2 =ax+3 for x=2 ⇒ a=(1/2)  but then it′s not continuous at 4 (it′s periodic ⇒  ⇒ it ♮starts againε at x=4 with the same  value as at x=0  f(0)=0  lim_(x→4^− ) f(x)=(1/2)×4+3=5  lim_(x→4^+ ) f(x)=0^2 =0  so it′s not continuous...  anyway the value of f(6)=f(6−4)=f(2)=4  closed dots at ≤ and ≥  open dots at < and >

$$\mathrm{if}\:{f}\left({x}\right)\:\mathrm{is}\:\mathrm{continuous}\:\mathrm{within}\:\left[\mathrm{0};\:\mathrm{4}\right]\:\Rightarrow \\ $$ $$\Rightarrow\:{x}^{\mathrm{2}} ={ax}+\mathrm{3}\:\mathrm{for}\:{x}=\mathrm{2}\:\Rightarrow\:{a}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$ $$\mathrm{but}\:\mathrm{then}\:\mathrm{it}'\mathrm{s}\:\mathrm{not}\:\mathrm{continuous}\:\mathrm{at}\:\mathrm{4}\:\left(\mathrm{it}'\mathrm{s}\:\mathrm{periodic}\:\Rightarrow\right. \\ $$ $$\Rightarrow\:\mathrm{it}\:\natural\mathrm{starts}\:\mathrm{again}\varepsilon\:\mathrm{at}\:{x}=\mathrm{4}\:\mathrm{with}\:\mathrm{the}\:\mathrm{same} \\ $$ $$\mathrm{value}\:\mathrm{as}\:\mathrm{at}\:{x}=\mathrm{0} \\ $$ $${f}\left(\mathrm{0}\right)=\mathrm{0} \\ $$ $$\underset{{x}\rightarrow\mathrm{4}^{−} } {\mathrm{lim}}{f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{4}+\mathrm{3}=\mathrm{5} \\ $$ $$\underset{{x}\rightarrow\mathrm{4}^{+} } {\mathrm{lim}}{f}\left({x}\right)=\mathrm{0}^{\mathrm{2}} =\mathrm{0} \\ $$ $$\mathrm{so}\:\mathrm{it}'\mathrm{s}\:\mathrm{not}\:\mathrm{continuous}... \\ $$ $$\mathrm{anyway}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{f}\left(\mathrm{6}\right)={f}\left(\mathrm{6}−\mathrm{4}\right)={f}\left(\mathrm{2}\right)=\mathrm{4} \\ $$ $$\mathrm{closed}\:\mathrm{dots}\:\mathrm{at}\:\leqslant\:\mathrm{and}\:\geqslant \\ $$ $$\mathrm{open}\:\mathrm{dots}\:\mathrm{at}\:<\:\mathrm{and}\:> \\ $$

Commented byRio Michael last updated on 14/Sep/19

thank you sir,so much

$${thank}\:{you}\:{sir},{so}\:{much} \\ $$

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