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Question Number 21070 by Tinkutara last updated on 11/Sep/17

$$\mathrm{Let}\mathrm{us}\mathrm{consider}\mathrm{an}\mathrm{equation}f\left(x\right)=x^3$$$$-3x+k=0.\mathrm{Then}\mathrm{the}\mathrm{values}\mathrm{of}k\mathrm{for}$$$$\mathrm{which}\mathrm{the}\mathrm{equation}\mathrm{has}$$$$1.\mathrm{Exactly}\mathrm{one}\mathrm{root}\mathrm{which}\mathrm{is}\mathrm{positive},$$$$\mathrm{then}k\mathrm{belongs}\mathrm{to}$$$$2.\mathrm{Exactly}\mathrm{one}\mathrm{root}\mathrm{which}\mathrm{is}\mathrm{negative},$$$$\mathrm{then}k\mathrm{belongs}\mathrm{to}$$$$3.\mathrm{One}\mathrm{negative}\mathrm{and}\mathrm{two}\mathrm{positive}\mathrm{root}$$$$\mathrm{if}k\mathrm{belongs}\mathrm{to}$$

Commented by Tinkutara last updated on 12/Sep/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!$$

Commented by mrW1 last updated on 12/Sep/17

$$1.\mathrm{k}\in (-\mathrm{\infty },-2)$$$$2.\mathrm{k}\in (2,+\mathrm{\infty })$$$$3.\mathrm{k}\in (0,2)$$

Answered by mrW1 last updated on 12/Sep/17

$${\mathrm{let}}^{\prime }\mathrm{s}\mathrm{look}\mathrm{at}\mathrm{the}\mathrm{case}\mathrm{k}=0:$$$$\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^3-3\mathrm{x}=0$$$$\mathrm{x}(\mathrm{x}-\sqrt{3})(\mathrm{x}+\sqrt{3})=0$$$$\mathrm{it}\mathrm{has}\mathrm{three}\mathrm{roots}:$$$$-\sqrt{3},0,\sqrt{3}$$$$\mathrm{f}(\mathrm{x}\to -\mathrm{\infty })\to -\mathrm{\infty }$$$$\mathrm{f}(\mathrm{x}\to +\mathrm{\infty })\to +\mathrm{\infty }$$$${\mathrm{f}}^{\prime }\left(\mathrm{x}\right)={3\mathrm{x}}^2-3=0$$$$\Rightarrow \mathrm{x}=-1,1$$$$\mathrm{f}(-1)=2=\mathrm{local}\max .$$$$\mathrm{f}\left(1\right)=-2=\mathrm{local}\min .$$$$$$$$\mathrm{the}\mathrm{graph}\mathrm{of}\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^3-3\mathrm{x}\mathrm{see}\mathrm{diagram}.$$

Commented by mrW1 last updated on 12/Sep/17

Commented by mrW1 last updated on 13/Sep/17

$$\mathrm{with}\mathrm{k}\ne 0\mathrm{the}\mathrm{graph}\mathrm{will}\mathrm{be}\mathrm{moved}\mathrm{up}$$$$(\mathrm{k}>0)\mathrm{or}\mathrm{down}(\mathrm{k}<0).$$$$1.\mathrm{so}\mathrm{that}\mathrm{there}\mathrm{is}\mathrm{only}\mathrm{one}+\mathrm{ve}\mathrm{root},$$$$\mathrm{the}\mathrm{graph}\mathrm{must}\mathrm{be}\mathrm{moved}\mathrm{down}\mathrm{at}\mathrm{least}$$$$\mathrm{by}2\mathrm{units},\mathrm{i}.\mathrm{e}.\mathrm{k}<-2$$$$2.\mathrm{so}\mathrm{that}\mathrm{there}\mathrm{is}\mathrm{only}\mathrm{one}-\mathrm{ve}\mathrm{root},$$$$\mathrm{the}\mathrm{graph}\mathrm{must}\mathrm{be}\mathrm{moved}\mathrm{up}\mathrm{at}\mathrm{least}$$$$\mathrm{by}2\mathrm{units},\mathrm{i}.\mathrm{e}.\mathrm{k}>2$$$$3.\mathrm{so}\mathrm{that}\mathrm{there}\mathrm{are}2+\mathrm{ve}\mathrm{roots}\mathrm{and}\mathrm{one}-\mathrm{ve}\mathrm{root},$$$$\mathrm{the}\mathrm{graph}\mathrm{must}\mathrm{be}\mathrm{moved}\mathrm{up},\mathrm{but}$$$$\mathrm{by}\mathrm{less}\mathrm{than}2\mathrm{units},\mathrm{i}.\mathrm{e}.0<\mathrm{k}<2.$$

Commented by Tinkutara last updated on 13/Sep/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!$$

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