Question Number 85425 by john santu last updated on 22/Mar/20
$${range}\:{of}\:{function} \\ $$$${y}\:=\:\frac{\mid{x}−\mathrm{1}\mid}{{x}+\mathrm{3}} \\ $$
Answered by john santu last updated on 22/Mar/20
$${for}\:{x}\:\geqslant\mathrm{1}\:\Rightarrow\:{y}\:=\:\frac{{x}−\mathrm{1}}{{x}+\mathrm{3}} \\ $$$${x}\:=\:\frac{−\mathrm{3}{y}−\mathrm{1}}{{y}−\mathrm{1}}=\:\frac{\mathrm{3}{y}+\mathrm{1}}{\mathrm{1}−{y}} \\ $$$${we}\:{get}\::\:\mathrm{0}\:\leqslant{y}\:<\mathrm{1}\: \\ $$$${for}\:{x}\:<\:\mathrm{1}\:\Rightarrow\:{y}\:=\:\frac{−{x}+\mathrm{1}}{{x}+\mathrm{3}} \\ $$$${x}\:=\:\frac{−\mathrm{3}{y}+\mathrm{1}}{{y}+\mathrm{1}}\:,\:{we}\:{get}\:{y}<−\mathrm{1}\:\vee\:{y}\geqslant\mathrm{0} \\ $$$${therefore}\:{range}\:{is}\:{y}\:<−\mathrm{1}\:\vee\:{y}\geqslant\:\mathrm{0} \\ $$$$ \\ $$
Commented by jagoll last updated on 22/Mar/20
Commented by jagoll last updated on 22/Mar/20
$$\mathrm{this}\:\mathrm{the}\:\mathrm{graph}\: \\ $$
Answered by MJS last updated on 22/Mar/20
$${y}=\begin{cases}{\frac{\mathrm{1}−{x}}{{x}+\mathrm{3}}=−\mathrm{1}+\frac{\mathrm{4}}{{x}+\mathrm{3}};\:{x}<\mathrm{1}}\\{\frac{{x}−\mathrm{1}}{{x}+\mathrm{3}}=\mathrm{1}−\frac{\mathrm{4}}{{x}+\mathrm{3}};\:{x}\geqslant\mathrm{1}}\end{cases} \\ $$$$\mathrm{range}=\begin{cases}{{R}_{\mathrm{1}} =\mathbb{R}\backslash\left[−\mathrm{1};\:\mathrm{0}\right]}\\{{R}_{\mathrm{2}} =\left[\mathrm{0};\:\mathrm{1}\left[\right.\right.}\end{cases} \\ $$$$\mathrm{range}:\:{R}={R}_{\mathrm{1}} \cup{R}_{\mathrm{2}} =\mathbb{R}\backslash\left[−\mathrm{1};\mathrm{0}\left[\right.\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\mathrm{or}\:{R}=\right]−\infty;\:−\mathrm{1}\left[\:\cup\:\left[\mathrm{0};\:+\infty\left[\right.\right.\right. \\ $$
Commented by john santu last updated on 22/Mar/20
$${my}\:{answer}\:{correct}\:{or}\:{wrong}? \\ $$
Commented by john santu last updated on 22/Mar/20
$${i}\:{don}'{t}\:{understand}\:{the} \\ $$$$\left.{notation}\:\right]−\infty;\:−\mathrm{1}\left[\:?\right. \\ $$$${it}\:{same}\:{to}\:\left(−\infty,−\mathrm{1}\right)\:?\:{sir} \\ $$
Commented by $@ty@m123 last updated on 22/Mar/20
$${Yeah}. \\ $$$${Both}\:{are}\:{same}. \\ $$