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Question Number 79279 by jagoll last updated on 24/Jan/20

$$\mathrm{solve}$$ $$\mid \mathrm{x}{\mid }^3-{7\mathrm{x}}^2+7\mid \mathrm{x}\mid +15<0$$

Commented byjagoll last updated on 24/Jan/20

$$\mathrm{thanks}$$

Commented byMJS last updated on 24/Jan/20

$$\mathrm{sorry}\mathrm{but}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{wrong}$$

Commented byjohn santu last updated on 24/Jan/20

$$whereiswrong?$$

Commented byMJS last updated on 24/Jan/20

$$\mathrm{you}\mathrm{should}\mathrm{plot}\mathrm{the}\mathrm{function}\mathrm{or}\mathrm{scetch}\mathrm{it}$$

Answered by MJS last updated on 24/Jan/20

$$\mathrm{case}1x⩾0$$ $$x^3-7x^2+7x+15=0$$ $$\Rightarrow x=-1\vee x=3\vee x=5$$ $$\mathrm{testing}x=4:4^3-7\times 4^2+7\times 4+15=-5$$ $$\mathrm{case}2x<0$$ $$-x^3-7x^2-7x+15=0$$ $$\Rightarrow x=-5\vee x=-3\vee x=1$$ $$\mathrm{testing}x=-4:-{(-4)}^3-7\times {(-4)}^2-7\times 4+15=-5$$ $$$$ $$\Rightarrow$$ $$-5<x<-3\vee 3<x<5$$ $$\mathrm{or}$$ $$3<\mid x\mid <5$$

Commented byjohn santu last updated on 24/Jan/20

$$yourareright.i^{\prime }mforgot$$ $$thedefinitionsign\left(x\right)$$

Commented byjagoll last updated on 24/Jan/20

$$\mathrm{thanks}\mathrm{you}\mathrm{sir}$$

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