Question Number 112464 by Aina Samuel Temidayo last updated on 08/Sep/20
Answered by john santu last updated on 08/Sep/20
$${f}\left({x}\right)=\mathrm{0} \\ $$$$\begin{cases}{\mathrm{2}^{{x}} \:=\:\mathrm{1}\:\rightarrow{x}=\mathrm{0}}\\{{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{3}=\mathrm{0}\rightarrow\left({x}+\mathrm{3}\right)\left({x}−\mathrm{1}\right)=\mathrm{0}}\end{cases} \\ $$$${x}=−\mathrm{3}\:;\:{x}\:=\:\mathrm{1} \\ $$
Commented by Aina Samuel Temidayo last updated on 08/Sep/20
$$\mathrm{Yea}. \\ $$