Mlejd5

2022-01-24

Find the number of roots of the equation, ${x}^{3}+{x}^{2}+2x+\mathrm{sin}x=0$ in $\left[-2\pi ,2\pi \right]$

Joy Compton

$\mathrm{\forall }x\left(f\prime \left(x\right)=3{x}^{2}+2x+2+\mathrm{cos}x>0\right)$ (the polynomial term is always at least $\frac{5}{3}$), so note that and conclude that there is exactly 1 root.

Gwendolyn Meyer

For x<0, $\mathrm{sin}x=x-\frac{{x}^{3}}{3!}+O\left({x}^{5}\right)>x:$
${x}^{3}+{x}^{2}+2x-\mathrm{sin}x<{x}^{3}+{x}^{2}+2x-x=x\left({x}^{2}+x+1\right)<0$,
for $x>0,\mathrm{sin}x:
${x}^{3}+{x}^{2}+2x-\mathrm{sin}x>{x}^{3}+{x}^{2}+2x-x={x}^{3}+{x}^{2}+x>0$

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