Question Number 14470 by Tinkutara last updated on 01/Jun/17
![Find the number of solution(s) of x^2 + x + sin x = 0, x ∈ [0, π]](Q14470.png)
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{solution}\left(\mathrm{s}\right)\:\mathrm{of} \\ $$$${x}^{\mathrm{2}} \:+\:{x}\:+\:\mathrm{sin}\:{x}\:=\:\mathrm{0},\:{x}\:\in\:\left[\mathrm{0},\:\pi\right] \\ $$
Commented by mrW1 last updated on 01/Jun/17
![since x ∈ [0, π] x^2 ≥0 x≥0 sin x≥0 x^2 +x+sin x≥0 “=” only at x=0 ⇒ x^2 + x + sin x = 0 has only one solution x=0](Q14471.png)
$${since}\:{x}\:\in\:\left[\mathrm{0},\:\pi\right] \\ $$$${x}^{\mathrm{2}} \geqslant\mathrm{0} \\ $$$${x}\geqslant\mathrm{0} \\ $$$$\mathrm{sin}\:{x}\geqslant\mathrm{0} \\ $$$${x}^{\mathrm{2}} +{x}+\mathrm{sin}\:{x}\geqslant\mathrm{0} \\ $$$$``=''\:{only}\:{at}\:{x}=\mathrm{0} \\ $$$$\Rightarrow\:{x}^{\mathrm{2}} \:+\:{x}\:+\:\mathrm{sin}\:{x}\:=\:\mathrm{0}\:{has}\:{only} \\ $$$${one}\:{solution}\:{x}=\mathrm{0} \\ $$
Commented by Tinkutara last updated on 01/Jun/17
$$\mathrm{But}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{of}\:{x}^{\mathrm{2}} \:+\:{x}\:\mathrm{is}\:\frac{−\mathrm{1}}{\mathrm{4}}\:\mathrm{at} \\ $$$${x}\:=\:\frac{−\mathrm{1}}{\mathrm{2}}\:.\:\mathrm{Will}\:\mathrm{it}\:\mathrm{not}\:\mathrm{affect}\:\mathrm{the}\:\mathrm{answer}? \\ $$
Commented by mrW1 last updated on 01/Jun/17
![x∈[0,π] so x≥0, and there is no minimum.](Q14475.png)
$${x}\in\left[\mathrm{0},\pi\right] \\ $$$${so}\:{x}\geqslant\mathrm{0},\:{and}\:{there}\:{is}\:{no}\:{minimum}. \\ $$
Commented by Tinkutara last updated on 01/Jun/17
$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$