Question Number 136138 by bramlexs22 last updated on 19/Mar/21
$${What}\:{is}\:{max}\:{and}\:{min}\:{value} \\ $$$${of}\:{y}=\mathrm{4}−{x}^{\mathrm{2}} −\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:? \\ $$
Answered by ajfour last updated on 19/Mar/21
$${let}\:\:\mathrm{1}−{x}^{\mathrm{2}} ={s}\geqslant\mathrm{0} \\ $$$$\Rightarrow\:\: \\ $$$${y}=\mathrm{3}+{s}−\sqrt{{s}} \\ $$$$\frac{{dy}}{{ds}}=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}\sqrt{{s}}}=\mathrm{0}\:\:\:\Rightarrow \\ $$$${s}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${y}=\mathrm{3}+\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{11}}{\mathrm{4}} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{ds}^{\mathrm{2}} }\mid_{{s}=\frac{\mathrm{1}}{\mathrm{4}}} =\left(\frac{\mathrm{1}}{\mathrm{4}{s}\sqrt{{s}}}\right)\mid_{{s}=\frac{\mathrm{1}}{\mathrm{4}}} >\mathrm{0} \\ $$$${hence}\:\:{y}_{{min}} =\frac{\mathrm{11}}{\mathrm{4}} \\ $$$${and}\:{for}\:{s}=\mathrm{0}\left({x}=\pm\mathrm{1}\right) \\ $$$${y}=\mathrm{3} \\ $$$${for}\:{s}=\mathrm{1}\left({x}=\mathrm{0}\right) \\ $$$${y}=\mathrm{3} \\ $$$${hence}\:{y}_{{max}} =\mathrm{3} \\ $$
Answered by mr W last updated on 19/Mar/21
$${y}=\mathrm{3}+\left(\mathrm{1}−{x}^{\mathrm{2}} \right)−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$${y}=\frac{\mathrm{11}}{\mathrm{4}}+\left(\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \geqslant\frac{\mathrm{11}}{\mathrm{4}} \\ $$$${y}_{{min}} =\frac{\mathrm{11}}{\mathrm{4}}\:{when}\:\mathrm{1}−{x}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}},\:{x}\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$${y}_{{max}} =\frac{\mathrm{11}}{\mathrm{4}}+\left(\pm\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{3}\:{when}\:{x}=\mathrm{0}\:{or}\:\pm\mathrm{1} \\ $$