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Question Number 147980 by mathdanisur last updated on 24/Jul/21

$$f\left(x\right)=4x^4-2x^2+17$$$$Findthemaximumpointofthe$$$$function$$

Answered by iloveisrael last updated on 25/Jul/21

$$\mathrm{f}{}^{\prime }\left(\mathrm{x}\right)={16\mathrm{x}}^3-4\mathrm{x}=0$$$$\Rightarrow 4\mathrm{x}({4\mathrm{x}}^2-1)=0$$$$\Rightarrow \mathrm{x}=0,\mathrm{x}=\pm \frac{1}{2}$$$$\mathrm{f}{}^{″}\left(\mathrm{x}\right)={48\mathrm{x}}^2-4<0\mathrm{for}\mathrm{x}=0$$$$\mathrm{local}\max \mathrm{when}\mathrm{x}=0,\mathrm{local}\max \mathrm{value}=17$$$$\mathrm{local}\max \mathrm{point}\mathrm{at}(0,17)$$$$$$

Commented by mathdanisur last updated on 25/Jul/21

$$ThanksSer$$

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