Question Number 39416 by Rio Mike last updated on 06/Jul/18
$${Given}\:{that}\:{f}\left({x}\right)\:{is}\:{a}\:{cubic}\:{function} \\ $$$${and}\:{f}\left({x}\right)\:=\:{x}^{\mathrm{3}} \:\:−\:\frac{{x}^{\mathrm{2}} }{\mathrm{4}}\:+\:\mathrm{5}{x}\:−\:\mathrm{7} \\ $$$$\left.{a}\right)\:{find}\:{one}\:{factor}\:{of}\:{f}\left({x}\right) \\ $$$$\left.{b}\right)\:{find}\:\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:{for}\:{f}\left({x}\right) \\ $$$$\left.{c}\right)\:{hence}\:{Evaluate}\:\:{y}\:=\:\int_{\mathrm{0}} ^{\infty} {f}\left({x}\right). \\ $$
Answered by MJS last updated on 06/Jul/18
![(b) (d^2 y/dx^2 )[f(x)]=f′′(x)=6x−(1/2) (c) ∫f(x)dx=(1/4)x^4 −(1/(12))x^3 +(5/2)x^2 −7x+C ⇒ ∫_0 ^∞ f(x)dx=+∞ (a) what are we allowed to use? approximation procedure with a calculator gives x≈1.15710 Cardano: x^3 −(1/4)x^2 +5x−7=0 z=x+(1/(12)) z^3 +((239)/(48))z−((5689)/(864))=0 u=(1/(12))((5689+24(√(79890))))^(1/3) ; v=(1/(12))((5689−24(√(79890))))^(1/3) z_1 =u+v; x_1 =(1/(12))+u+v≈1.15710 z_2 =(−(1/2)+((√3)/2)i)u+(−(1/2)−((√3)/2)i)v z_2 =(−(1/2)−((√3)/2)i)u+(−(1/2)+((√3)/2)i)v](Q39423.png)
$$\left({b}\right)\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\left[{f}\left({x}\right)\right]={f}''\left({x}\right)=\mathrm{6}{x}−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\left({c}\right)\:\int{f}\left({x}\right){dx}=\frac{\mathrm{1}}{\mathrm{4}}{x}^{\mathrm{4}} −\frac{\mathrm{1}}{\mathrm{12}}{x}^{\mathrm{3}} +\frac{\mathrm{5}}{\mathrm{2}}{x}^{\mathrm{2}} −\mathrm{7}{x}+{C}\:\Rightarrow\: \\ $$$$\:\:\:\:\:\underset{\mathrm{0}} {\overset{\infty} {\int}}{f}\left({x}\right){dx}=+\infty \\ $$$$\left({a}\right)\:\mathrm{what}\:\mathrm{are}\:\mathrm{we}\:\mathrm{allowed}\:\mathrm{to}\:\mathrm{use}? \\ $$$$\:\:\:\:\:\mathrm{approximation}\:\mathrm{procedure}\:\mathrm{with}\:\mathrm{a}\:\mathrm{calculator} \\ $$$$\:\:\:\:\:\mathrm{gives}\:{x}\approx\mathrm{1}.\mathrm{15710} \\ $$$$\:\:\:\:\:\mathrm{Cardano}: \\ $$$$\:\:\:\:\:{x}^{\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{4}}{x}^{\mathrm{2}} +\mathrm{5}{x}−\mathrm{7}=\mathrm{0} \\ $$$$\:\:\:\:\:{z}={x}+\frac{\mathrm{1}}{\mathrm{12}} \\ $$$$\:\:\:\:\:{z}^{\mathrm{3}} +\frac{\mathrm{239}}{\mathrm{48}}{z}−\frac{\mathrm{5689}}{\mathrm{864}}=\mathrm{0} \\ $$$$\:\:\:\:\:{u}=\frac{\mathrm{1}}{\mathrm{12}}\sqrt[{\mathrm{3}}]{\mathrm{5689}+\mathrm{24}\sqrt{\mathrm{79890}}};\:{v}=\frac{\mathrm{1}}{\mathrm{12}}\sqrt[{\mathrm{3}}]{\mathrm{5689}−\mathrm{24}\sqrt{\mathrm{79890}}} \\ $$$$\:\:\:\:\:{z}_{\mathrm{1}} ={u}+{v};\:{x}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{12}}+{u}+{v}\approx\mathrm{1}.\mathrm{15710} \\ $$$$\:\:\:\:\:{z}_{\mathrm{2}} =\left(−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right){u}+\left(−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right){v} \\ $$$$\:\:\:\:\:{z}_{\mathrm{2}} =\left(−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right){u}+\left(−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right){v} \\ $$