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Question Number 66149 by Rio Michael last updated on 09/Aug/19

$$f\left(x\right)=2x^3-x-4$$$$showthatf\left(x\right)=0hasrootsbetween$$$$1and2$$

Answered by MJS last updated on 09/Aug/19

$$f\left(1\right)=-3<0$$$$f\left(2\right)=10>0$$$$\mathrm{and}f\left(x\right)\mathrm{is}\mathrm{continuous}\mathrm{for}x\in \mathbb{R}$$$$\Rightarrow \mathrm{there}\mathrm{nust}\mathrm{be}\mathrm{at}\mathrm{least}\mathrm{one}\mathrm{zero}\mathrm{between}$$$$x=1\mathrm{and}x=2$$

Answered by mr W last updated on 09/Aug/19

$$f^{\prime }\left(x\right)=6x^2-1=0\Rightarrow x=\pm \frac{1}{\sqrt{6}}$$$$i.e.forx>\frac{1}{\sqrt{6}}wehave{f}^{\prime }\left(x\right)>0\Rightarrow strictlyincreasing$$$$f\left(1\right)=2-1-4=-3<0$$$$f\left(2\right)=16-2-4=10>0$$$$\Rightarrow f\left(x\right)=0hasoneandonlyoneroot$$$$between1and2.$$

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