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Question Number 39416 by Rio Mike last updated on 06/Jul/18

$$Giventhatf\left(x\right)isacubicfunction$$$$andf\left(x\right)=x^3-\frac{x^2}{4}+5x-7$$$$a)findonefactoroff\left(x\right)$$$$b)find\frac{d^{2}y}{{dx}^2}forf\left(x\right)$$$$c)henceEvaluatey={\int }_0^{\mathrm{\infty }}f\left(x\right).$$

Answered by MJS last updated on 06/Jul/18

$$\left(b\right)\frac{d^{2}y}{{dx}^2}\left[f\left(x\right)\right]=f^{″}\left(x\right)=6x-\frac{1}{2}$$$$\left(c\right)\int f\left(x\right)dx=\frac{1}{4}{x}^4-\frac{1}{12}{x}^3+\frac{5}{2}{x}^2-7x+C\Rightarrow$$$$\underset{0}{\stackrel{\mathrm{\infty }}{\int }}f\left(x\right)dx=+\mathrm{\infty }$$$$\left(a\right)\mathrm{what}\mathrm{are}\mathrm{we}\mathrm{allowed}\mathrm{to}\mathrm{use}?$$$$\mathrm{approximation}\mathrm{procedure}\mathrm{with}\mathrm{a}\mathrm{calculator}$$$$\mathrm{gives}x\approx 1.15710$$$$\mathrm{Cardano}:$$$$x^3-\frac{1}{4}{x}^2+5x-7=0$$$$z=x+\frac{1}{12}$$$$z^3+\frac{239}{48}z-\frac{5689}{864}=0$$$$u=\frac{1}{12}\sqrt[3]{5689+24\sqrt{79890}};v=\frac{1}{12}\sqrt[3]{5689-24\sqrt{79890}}$$$$z_1=u+v;x_1=\frac{1}{12}+u+v\approx 1.15710$$$$z_2=(-\frac{1}{2}+\frac{\sqrt{3}}{2}\mathrm{i})u+(-\frac{1}{2}-\frac{\sqrt{3}}{2}\mathrm{i})v$$$$z_2=(-\frac{1}{2}-\frac{\sqrt{3}}{2}\mathrm{i})u+(-\frac{1}{2}+\frac{\sqrt{3}}{2}\mathrm{i})v$$

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