Question Number 124404 by benjo_mathlover last updated on 03/Dec/20
$${Find}\:{minimum}\:{value}\:{of}\:{function} \\ $$$${f}\left({x}\right)=\mathrm{sin}\:^{\mathrm{7}} \left({x}\right)+\mathrm{cos}\:^{\mathrm{11}} \left({x}\right) \\ $$
Commented by benjo_mathlover last updated on 03/Dec/20
Commented by benjo_mathlover last updated on 03/Dec/20
$${yes}.\:{but}\:{how}\:{the}\:{way}\:{in}\:{algebraic}?? \\ $$
Commented by MJS_new last updated on 03/Dec/20
$$−\mathrm{1}\leqslant{f}\left({x}\right)\leqslant\mathrm{1} \\ $$$$\mathrm{is}\:\mathrm{there}\:\mathrm{an}\:\mathrm{easy}\:\mathrm{way}\:\mathrm{to}\:\mathrm{show}? \\ $$
Answered by mindispower last updated on 03/Dec/20
$${sin}^{\mathrm{7}} \left({x}\right)={sin}^{\mathrm{2}} \left({x}\right).{sin}^{\mathrm{5}} \left({x}\right) \\ $$$$−\mathrm{1}\leqslant{sin}^{\mathrm{5}} \left({x}\right)\leqslant\mathrm{1}, \\ $$$$−\mathrm{1}\leqslant{cos}^{\mathrm{9}} \left({x}\leqslant\mathrm{1}\right. \\ $$$$\Rightarrow−{sin}^{\mathrm{2}} \left({x}\right)−{cos}^{\mathrm{2}} \left({x}\right)\leqslant{sin}^{\mathrm{7}} \left({x}\right)+{cos}^{\mathrm{11}} \left({x}\right)\leqslant{sin}^{\mathrm{2}} \left({x}\right)+{cos}^{\mathrm{2}} \left({x}\right) \\ $$$$\Leftrightarrow \\ $$$$−\mathrm{1}\leqslant{f}\left({x}\right)\leqslant\mathrm{1} \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{1} \\ $$$${f}\left(−\frac{\pi}{\mathrm{2}}\right)=−\mathrm{1} \\ $$$${minf}=−\mathrm{1},{maxf}=\mathrm{1} \\ $$