Find the number of roots of the equation, x3+x2+2x+sinx=0 in [−2π,2π]

Mlejd5

Answered question

2022-01-24

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

Answer & Explanation

Joy Compton

Beginner2022-01-25Added 13 answers

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

Gwendolyn Meyer

Beginner2022-01-26Added 11 answers

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