Question Number 44351 by Necxx last updated on 27/Sep/18
$${If}\:{is}\:{an}\:{even}\:{function}\:{defined}\:{on} \\ $$$${the}\:{interval}\:\left(−\mathrm{5},\mathrm{5}\right)\:{then}\:{a}\:{value} \\ $$$${of}\:{x}\:{satisfying}\:{the}\:{equation} \\ $$$${f}\left({x}\right)={f}\left(\frac{{x}+\mathrm{1}}{{x}+\mathrm{2}}\right)\:{is} \\ $$$$\left.{a}\left.\right)\left.\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:{b}\right)\frac{−\mathrm{3}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:{c}\right)\frac{−\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\: \\ $$$$\left.{d}\right)\frac{−\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{2}_{} } \\ $$
Commented by Necxx last updated on 27/Sep/18
$${please}\:{help} \\ $$
Answered by MrW3 last updated on 27/Sep/18
$${f}\left({x}\right)={f}\left(−{x}\right) \\ $$$${f}\left({x}\right)={f}\left(\frac{{x}+\mathrm{1}}{{x}+\mathrm{2}}\right) \\ $$$$ \\ $$$${case}\:\mathrm{1}:\:\frac{{x}+\mathrm{1}}{{x}+\mathrm{2}}={x} \\ $$$${x}+\mathrm{1}={x}^{\mathrm{2}} +\mathrm{2}{x} \\ $$$${x}^{\mathrm{2}} +{x}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{x}=\frac{−\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$ \\ $$$${case}\:\mathrm{2}:\:\frac{{x}+\mathrm{1}}{{x}+\mathrm{2}}=−{x} \\ $$$${x}+\mathrm{1}=−{x}^{\mathrm{2}} −\mathrm{2}{x} \\ $$$${x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{x}=\frac{−\mathrm{3}\pm\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$ \\ $$$$\left.{a}\left.\right)\left.{b}\left.\right){c}\right){d}\right)\:{are}\:{all}\:{correct}. \\ $$
Commented by Necxx last updated on 27/Sep/18
$${Thank}\:{you}\:{sir}.{I}\:{appreciate}. \\ $$