Question Number 23814 by tawa tawa last updated on 06/Nov/17
Commented by tawa tawa last updated on 06/Nov/17
$$\mathrm{please}\:\mathrm{help}. \\ $$
Answered by mrW1 last updated on 06/Nov/17
$$\Rightarrow{x}\geqslant−\frac{{a}}{{b}} \\ $$$$\Rightarrow{x}\geqslant−\frac{{b}}{{c}} \\ $$$$\Rightarrow{x}\geqslant−\frac{{c}}{{a}} \\ $$$$\Rightarrow{x}^{\mathrm{3}} \geqslant−\mathrm{1} \\ $$$$\Rightarrow{x}\geqslant−\mathrm{1} \\ $$$$ \\ $$$$\Rightarrow{x}\leqslant\frac{{b}}{{a}} \\ $$$$\Rightarrow{x}\leqslant\frac{{c}}{{b}} \\ $$$$\Rightarrow{x}\leqslant\frac{{a}}{{c}} \\ $$$$\Rightarrow{x}^{\mathrm{3}} \leqslant\mathrm{1} \\ $$$$\Rightarrow{x}\leqslant\mathrm{1} \\ $$$$ \\ $$$${x}=\mathrm{0}\:{is}\:{the}\:{solution}. \\ $$
Commented by tawa tawa last updated on 06/Nov/17
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Commented by Joel577 last updated on 07/Nov/17
$$\mathrm{If}\:{x}\:\geqslant\:−\mathrm{1}\:\mathrm{and}\:{x}\:\leqslant\:\mathrm{1},\:\mathrm{why}\:{x}\:=\:\mathrm{0}\:\mathrm{is}\:\mathrm{only}\:\mathrm{the}\:\mathrm{solution}? \\ $$
Commented by mrW1 last updated on 07/Nov/17
$${for}\:{the}\:{solution}\:{we}\:{don}'{t}\:{need}\:{to}\:{know} \\ $$$${that}\:−\mathrm{1}\leqslant{x}\leqslant\mathrm{1}. \\ $$$${we}\:{know}\:{that}\:{x}=\mathrm{0}\:{is}\:{a}\:{solution}. \\ $$$${we}\:{know}\:{theLHS}\:{is}\:{strictly}\:{increasing} \\ $$$${and}\:{the}\:{RHS}\:{is}\:{strictly}\:{decreasing}. \\ $$$${that}\:{means}\:{if}\:{there}\:{is}\:{a}\:{solution}\:{then} \\ $$$${this}\:{solution}\:{is}\:{the}\:{only}\:{solution}. \\ $$
Commented by Joel577 last updated on 07/Nov/17
$${understood},\:{Sir}.\:{Thank}\:{you} \\ $$