Question Number 67684 by Rio Michael last updated on 30/Aug/19
$${given}\:{the}\:{function}\: \\ $$$${f}\left({x}\right)\:=\begin{cases}{{x}^{\mathrm{2}} \:\:,\:{for}\:\:\:\mathrm{0}\leqslant\:{x}<\:\mathrm{2}}\\{{ax}\:+\:\mathrm{3},\:{for}\:\:\mathrm{2}\leqslant\:{x}\:<\:\mathrm{4}}\end{cases} \\ $$$${is}\:{periodic}\:{of}\:{period}\:\:\mathrm{4},\:{and}\:{is}\:{continuous}. \\ $$$$\left.{a}\right)\:{Find}\:\:{the}\:{value}\:{of}\:\:{a}. \\ $$$$\left.{b}\right)\:{Find}\:{the}\:{valu}\:{of}\:\:{f}\left(\mathrm{6}\right) \\ $$$$\left.{c}\right)\:{sketch}\:{the}\:{graph}\:{for}\:{y}\:={f}\left({x}\right). \\ $$$${help}\:{me}\:{please},\:{for}\:{the}\:{graph}\:{i}\:{don}'{t}\:{know}\:{wbere}\:{to}\:{put}\:\:{y}={x}^{\mathrm{2}} \:{and}\:{y}\:=\:{ax}\:+\:\mathrm{3}\:{and} \\ $$$${where}\:{do}\:{i}\:{put}\:{a}\:{closed}\:\:{dot}\:{and}\:{an}\:{open}\:{dot}. \\ $$$$ \\ $$
Commented by MJS 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 again” 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 >](https://www.tinkutara.com/question/Q67712.png)
$$\mathrm{if}\:{f}\left({x}\right)\:\mathrm{is}\:\mathrm{continuous}\:\mathrm{within}\:\left[\mathrm{0};\:\mathrm{4}\right]\:\Rightarrow \\ $$$$\Rightarrow\:{x}^{\mathrm{2}} ={ax}+\mathrm{3}\:\mathrm{for}\:{x}=\mathrm{2}\:\Rightarrow\:{a}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{but}\:\mathrm{then}\:\mathrm{it}'\mathrm{s}\:\mathrm{not}\:\mathrm{continuous}\:\mathrm{at}\:\mathrm{4}\:\left(\mathrm{it}'\mathrm{s}\:\mathrm{periodic}\:\Rightarrow\right. \\ $$$$\Rightarrow\:\mathrm{it}\:“\mathrm{starts}\:\mathrm{again}''\:\mathrm{at}\:{x}=\mathrm{4}\:\mathrm{with}\:\mathrm{the}\:\mathrm{same} \\ $$$$\mathrm{value}\:\mathrm{as}\:\mathrm{at}\:{x}=\mathrm{0} \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$$\underset{{x}\rightarrow\mathrm{4}^{−} } {\mathrm{lim}}{f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{4}+\mathrm{3}=\mathrm{5} \\ $$$$\underset{{x}\rightarrow\mathrm{4}^{+} } {\mathrm{lim}}{f}\left({x}\right)=\mathrm{0}^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{so}\:\mathrm{it}'\mathrm{s}\:\mathrm{not}\:\mathrm{continuous}… \\ $$$$\mathrm{anyway}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{f}\left(\mathrm{6}\right)={f}\left(\mathrm{6}−\mathrm{4}\right)={f}\left(\mathrm{2}\right)=\mathrm{4} \\ $$$$\mathrm{closed}\:\mathrm{dots}\:\mathrm{at}\:\leqslant\:\mathrm{and}\:\geqslant \\ $$$$\mathrm{open}\:\mathrm{dots}\:\mathrm{at}\:<\:\mathrm{and}\:> \\ $$
Commented by Rio Michael last updated on 14/Sep/19
$${thank}\:{you}\:{sir},{so}\:{much} \\ $$