Question Number 146253 by mathdanisur last updated on 12/Jul/21
$$\left(\mathrm{4}{x}^{\mathrm{4}} \:-\:\mathrm{2}{x}^{\mathrm{2}} \:+\:\mathrm{17}\right)_{\boldsymbol{{max}}} \:=\:? \\ $$
Answered by iloveisrael last updated on 12/Jul/21
$$\mathrm{f}\left(\mathrm{x}\right)_{\mathrm{min}} =\:\mathrm{4}\left(\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} −\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{4}}\right)+\mathrm{17} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)_{\mathrm{min}} \:=\:\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{2}}{\mathrm{4}}+\frac{\mathrm{68}}{\mathrm{4}}=\frac{\mathrm{67}}{\mathrm{4}}\:,\:\mathrm{when}\:\mathrm{x}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)_{\mathrm{max}} =\infty\: \\ $$
Commented by mathdanisur last updated on 12/Jul/21
$${thankyou}\:{Ser} \\ $$
Answered by gsk2684 last updated on 12/Jul/21
$$\mathrm{4}\left({x}^{\mathrm{4}} −\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} \right)+\mathrm{17} \\ $$$$\mathrm{4}\left(\left({x}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{16}}\right)+\mathrm{17} \\ $$$$\mathrm{4}\left({x}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}+\mathrm{17} \\ $$$$\mathrm{4}\left({x}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} +\frac{\mathrm{67}}{\mathrm{4}} \\ $$$${minimum}\:{is}\:\frac{\mathrm{67}}{\mathrm{4}} \\ $$$${maximum}\:{can}\:{not}\:{define}\:{on}\:{R} \\ $$
Commented by mathdanisur last updated on 12/Jul/21
$${thankyou}\:{Ser} \\ $$