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

∀x(f′(x)=3x2+2x+2+cos⁡x&gt;0) (the polynomial term is always at least 53), so note that f(−2π)&lt;0, f(2π)&gt;0 and conclude that there is exactly 1 root.

For x&lt;0, sin⁡x=x−x33!+O(x5)&gt;x: x3+x2+2x−sin⁡x&lt;x3+x2+2x−x=x(x2+x+1)&lt;0, for x&gt;0,sin⁡x&lt;x: x3+x2+2x−sin⁡x&gt;x3+x2+2x−x=x3+x2+x&gt;0

∀x(f′(x)=3x2+2x+2+cos⁡x&gt;0) (the polynomial term is always at least 53), so note that f(−2π)&lt;0, f(2π)&gt;0 and conclude that there is exactly 1 root.

For x&lt;0, sin⁡x=x−x33!+O(x5)&gt;x: x3+x2+2x−sin⁡x&lt;x3+x2+2x−x=x(x2+x+1)&lt;0, for x&gt;0,sin⁡x&lt;x: x3+x2+2x−sin⁡x&gt;x3+x2+2x−x=x3+x2+x&gt;0

Explained: "Find the number of roots of the equation,..."

Mlejd5

Answered question

2022-01-24

Find the number of roots of the equation,

x^{3} + x^{2} + 2 x + \sin x = 0

[- 2 π, 2 π]

Answer & Explanation

Joy Compton

Beginner2022-01-25Added 13 answers

\forall x (f' (x) = 3 x^{2} + 2 x + 2 + \cos x > 0)

(the polynomial term is always at least

\frac{5}{3}

), so note that

f (- 2 π) < 0, f (2 π) > 0

and conclude that there is exactly 1 root.

Gwendolyn Meyer

Beginner2022-01-26Added 11 answers

For x<0,

\sin x = x - \frac{x^{3}}{3!} + O (x^{5}) > x :

x^{3} + x^{2} + 2 x - \sin x < x^{3} + x^{2} + 2 x - x = x (x^{2} + x + 1) < 0

,
for

x > 0, \sin x < x

x^{3} + x^{2} + 2 x - \sin x > x^{3} + x^{2} + 2 x - x = x^{3} + x^{2} + x > 0

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