Question Number 169319 by otchereabdullai@gmail.com last updated on 28/Apr/22
$$\mathrm{if}\:\mathrm{f}\left(\mathrm{2x}−\mathrm{1}\right)=\mathrm{x}^{\mathrm{2}} −\mathrm{3x}+\mathrm{2},\:\:\mathrm{find}\:\mathrm{f}\left(\mathrm{2}\right) \\ $$
Commented by infinityaction last updated on 28/Apr/22
$${x}\rightarrow\frac{{x}+\mathrm{1}}{\mathrm{2}} \\ $$$${f}\left({x}\right)\:=\:\left(\frac{{x}+\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}\:} −\mathrm{3}\left(\frac{{x}+\mathrm{1}}{\mathrm{2}}\right)+\mathrm{2} \\ $$$${f}\left(\mathrm{2}\right)\:=\:\frac{\mathrm{9}}{\mathrm{4}}\:−\frac{\mathrm{9}}{\mathrm{2}}+\mathrm{2} \\ $$$${f}\left(\mathrm{2}\right)\:=\:−\frac{\mathrm{1}}{\mathrm{4}} \\ $$
Commented by otchereabdullai@gmail.com last updated on 28/Apr/22
$$\mathrm{Thank}\:\mathrm{you} \\ $$
Answered by alephzero last updated on 28/Apr/22
$${f}\left(\mathrm{2}{x}−\mathrm{1}\right)\:=\:{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{2} \\ $$$$\mathrm{2}\:=\:\mathrm{2}{x}−\mathrm{1} \\ $$$$\Rightarrow\:{x}\:=\:\mathrm{1}.\mathrm{5} \\ $$$$\Rightarrow\:{f}\left(\mathrm{2}\left(\mathrm{1}.\mathrm{5}\right)−\mathrm{1}\right)\:=\:\mathrm{1}.\mathrm{5}^{\mathrm{2}} −\mathrm{3}\left(\mathrm{1}.\mathrm{5}\right)+\mathrm{2} \\ $$$$=\:\mathrm{2}.\mathrm{25}−\mathrm{4}.\mathrm{5}+\mathrm{2}\:=\:−\mathrm{0}.\mathrm{25} \\ $$
Commented by otchereabdullai@gmail.com last updated on 28/Apr/22
$$\mathrm{thank}\:\mathrm{you} \\ $$