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4x-4-2x-2-17-max-




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

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