Question-219281

Question Number 219281 by Rojarani last updated on 22/Apr/25

Answered by Ghisom last updated on 22/Apr/25

P(x)=x^3 −3x+1 P ′(x)=3x^2 −3 P ′′(x)=6x x→−∞ P(x)→−∞ −∞<x<−1 P ′(x)>0 ⇒ P(x) increasing x=−1 P ′(x)=0∧P ′′(x)<0 ⇒ maximum (((−1)),(3) ) −1<x<1 P ′(x)<0 ⇒ P(x) decreasing x=1 P ′(x)=∧P ′′(x)>0 ⇒ minimum ((1),((−1)) ) 1<x<∞ P ′(x)>0 ⇒ P(x) increasing x→∞ P(x)→∞ ⇒ there must each one zero in 1.x∈(−∞, −1) 2. x∈(−1, 1) 3. x∈(1, ∞) using the trigonometric solution method we get a=−2cos (π/9) ≈−1.88 b=2sin (π/(18)) ≈.347 c=2cos ((2π)/9) ≈1.53 c−2bc=2−2cos ((2π)/9) ≈.468

$${P}\left({x}\right)={x}^{\mathrm{3}} −\mathrm{3}{x}+\mathrm{1} \\ $$$${P}\:'\left({x}\right)=\mathrm{3}{x}^{\mathrm{2}} −\mathrm{3} \\ $$$${P}\:''\left({x}\right)=\mathrm{6}{x} \\ $$$$ \\ $$$${x}\rightarrow−\infty \\ $$$$\:\:\:\:\:{P}\left({x}\right)\rightarrow−\infty \\ $$$$−\infty<{x}<−\mathrm{1} \\ $$$$\:\:\:\:\:{P}\:'\left({x}\right)>\mathrm{0}\:\Rightarrow\:{P}\left({x}\right)\:\mathrm{increasing} \\ $$$${x}=−\mathrm{1} \\ $$$$\:\:\:\:\:{P}\:'\left({x}\right)=\mathrm{0}\wedge{P}\:''\left({x}\right)<\mathrm{0}\:\Rightarrow\:\mathrm{maximum}\:\begin{pmatrix}{−\mathrm{1}}\\{\mathrm{3}}\end{pmatrix} \\ $$$$−\mathrm{1}<{x}<\mathrm{1} \\ $$$$\:\:\:\:\:{P}\:'\left({x}\right)<\mathrm{0}\:\Rightarrow\:{P}\left({x}\right)\:\mathrm{decreasing} \\ $$$${x}=\mathrm{1} \\ $$$$\:\:\:\:\:{P}\:'\left({x}\right)=\wedge{P}\:''\left({x}\right)>\mathrm{0}\:\Rightarrow\:\mathrm{minimum}\:\begin{pmatrix}{\mathrm{1}}\\{−\mathrm{1}}\end{pmatrix} \\ $$$$\mathrm{1}<{x}<\infty \\ $$$$\:\:\:\:\:{P}\:'\left({x}\right)>\mathrm{0}\:\Rightarrow\:{P}\left({x}\right)\:\mathrm{increasing} \\ $$$${x}\rightarrow\infty \\ $$$$\:\:\:\:\:{P}\left({x}\right)\rightarrow\infty \\ $$$$\Rightarrow\:\mathrm{there}\:\mathrm{must}\:\mathrm{each}\:\mathrm{one}\:\mathrm{zero}\:\mathrm{in} \\ $$$$\:\:\:\:\:\mathrm{1}.{x}\in\left(−\infty,\:−\mathrm{1}\right) \\ $$$$\:\:\:\:\:\mathrm{2}.\:{x}\in\left(−\mathrm{1},\:\mathrm{1}\right) \\ $$$$\:\:\:\:\:\mathrm{3}.\:{x}\in\left(\mathrm{1},\:\infty\right) \\ $$$$\mathrm{using}\:\mathrm{the}\:\mathrm{trigonometric}\:\mathrm{solution}\:\mathrm{method} \\ $$$$\mathrm{we}\:\mathrm{get} \\ $$$${a}=−\mathrm{2cos}\:\frac{\pi}{\mathrm{9}}\:\approx−\mathrm{1}.\mathrm{88} \\ $$$${b}=\mathrm{2sin}\:\frac{\pi}{\mathrm{18}}\:\approx.\mathrm{347} \\ $$$${c}=\mathrm{2cos}\:\frac{\mathrm{2}\pi}{\mathrm{9}}\:\approx\mathrm{1}.\mathrm{53} \\ $$$${c}−\mathrm{2}{bc}=\mathrm{2}−\mathrm{2cos}\:\frac{\mathrm{2}\pi}{\mathrm{9}}\:\approx.\mathrm{468} \\ $$

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