Question and Answers Forum

All Questions      Topic List

Limits Questions

Previous in All Question      Next in All Question      

Previous in Limits      Next in Limits      

Question Number 125815 by Tanuidesire last updated on 14/Dec/20

$$provethatthefunctionf\left(x\right)=\frac{1}{e^x-2}-2x,hasarootintheopeninterval(0,1).recallthate\approx 2.7$$

Answered by physicstutes last updated on 14/Dec/20

$$f\left(x\right)=\frac{1}{e^x-2}-2x$$$$f\left(0\right)=\frac{1}{-1}-0=-1$$$$f\left(1\right)=\frac{1}{e-2}-2\approx -0.6$$$$f\left(1\right)\times f\left(0\right)>0\mathrm{hence}f\left(x\right)\mathrm{does}\mathrm{not}\mathrm{have}\mathrm{roots}\mathrm{in}\mathrm{that}\mathrm{inteva}$$

Commented by Tanuidesire last updated on 14/Dec/20

$$thanks!$$

Commented by mindispower last updated on 14/Dec/20

$$f\left(0\right).f\left(1\right)<0withecontnuity\Rightarrow f\left(x\right)=0hasesolution$$$$f\left(0\right).f\left(1\right)>0wecansaynothing$$$$exempl$$$$f\left(x\right)=x+1,f\left(0\right).f\left(1\right)=2>0f\left(x\right)=0hasenosolution$$$$f\left(x\right)={(x-\frac{1}{2})}^2$$$$f\left(0\right)f\left(1\right)=\frac{1}{16}>0f\left(x\right)=0hasesolutionx=\frac{1}{2}$$$$$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com