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Question Number 146253 by mathdanisur last updated on 12/Jul/21

$${(4x^4-2x^2+17)}_{\mathit{m}\mathit{a}\mathit{x}}=?$$

Answered by iloveisrael last updated on 12/Jul/21

$$\mathrm{f}{\left(\mathrm{x}\right)}_{\min }=4{\left(\frac{1}{4}\right)}^2-2\left(\frac{1}{4}\right)+17$$$$\mathrm{f}{\left(\mathrm{x}\right)}_{\min }=\frac{1}{4}-\frac{2}{4}+\frac{68}{4}=\frac{67}{4},\mathrm{when}{\mathrm{x}}^2=\frac{1}{4}$$$$\mathrm{f}{\left(\mathrm{x}\right)}_{\max }=\mathrm{\infty }$$

Commented by mathdanisur last updated on 12/Jul/21

$$thankyouSer$$

Answered by gsk2684 last updated on 12/Jul/21

$$4(x^4-\frac{1}{2}{x}^2)+17$$$$4({(x^2-\frac{1}{4})}^2-\frac{1}{16})+17$$$$4{(x^2-\frac{1}{4})}^2-\frac{1}{4}+17$$$$4{(x^2-\frac{1}{4})}^2+\frac{67}{4}$$$$minimumis\frac{67}{4}$$$$maximumcannotdefineonR$$

Commented by mathdanisur last updated on 12/Jul/21

$$thankyouSer$$

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