Question Number 166443 by mathls last updated on 20/Feb/22 | ||
$${fog}_{\left(\mathrm{3}\right)} =\mathrm{10} \\ $$$${f}\left(\mathrm{3}\right)=\mathrm{4} \\ $$$${g}\left({x}\right)=? \\ $$ | ||
Commented by mr W last updated on 20/Feb/22 | ||
$${no}\:{solution}! \\ $$$${this}\:{question}\:{is}\:{something}\:{like}\:{the} \\ $$$${question}:\:{given}\:{that}\:{you}\:{go}\:{to}\:{the}\: \\ $$$${second}\:{class},\:{find}\:{the}\:{name}\:{of}\:{your}\: \\ $$$${grandpa}. \\ $$ | ||
Answered by ziaullah_Azizi last updated on 22/Feb/22 | ||
$$ \\ $$ | ||
Commented by ziaullah_Azizi last updated on 22/Feb/22 | ||
Commented by mr W last updated on 22/Feb/22 | ||
$${f}\left(\mathrm{3}\right)=\mathrm{4}\:\Rightarrow{f}\left({x}\right)={x}+\mathrm{1}\:???? \\ $$$${why}\:{not}\:{f}\left({x}\right)=\mathrm{2}\left({x}−\mathrm{1}\right)\:{or}\:{f}\left({x}\right)={x}^{\mathrm{2}} −\mathrm{5}\: \\ $$$${or}\:{f}\left({x}\right)=\mathrm{2}^{{x}} −\mathrm{4}\:{or}...? \\ $$ | ||
Answered by ziaullah_Azizi last updated on 22/Feb/22 | ||
$$ \\ $$ | ||