Question and Answers Forum

All Questions      Topic List

Relation and Functions Questions

Previous in All Question      Next in All Question      

Previous in Relation and Functions      Next in Relation and Functions      

Question Number 11608 by agni5 last updated on 29/Mar/17

$$\mathrm{If}\mathrm{f}\left(\mathrm{x}\right)={\mathrm{xtan}}^{-1}\left(\frac{1}{\mathrm{x}}\right),\mathrm{x}\ne 0$$$$=0,\mathrm{x}=0$$$$\mathrm{show}\mathrm{that}\mathrm{f}\mathrm{is}\mathrm{countinous}\mathrm{but}\mathrm{not}\mathrm{differentiable}$$$$\mathrm{at}\mathrm{x}=0.$$

Answered by mrW1 last updated on 30/Mar/17

$$f\left(0\right)=0$$$$\underset{x\to 0}{\lim }f\left(x\right)=\underset{x\to 0}{\lim }x{\tan }^{-1}\left(\frac{1}{x}\right)=\left(\underset{x\to 0}{\lim }x\right)\times \left(\underset{x\to 0}{\lim }{\tan }^{-1}\frac{1}{x}\right)$$$$=0\times (\pm \frac{\pi }{2})=0$$$$sincef\left(0\right)=\underset{x\to 0}{\lim }f\left(x\right)$$$$\Rightarrow f\left(x\right)iscontinousatx=0.$$$$$$$$f^{\prime }\left(x\right)={\tan }^{-1}\left(\frac{1}{x}\right)+x\left(\frac{1}{1+\frac{1}{x^2}}\right)(-\frac{1}{x^2})={\tan }^{-1}\left(\frac{1}{x}\right)-\frac{x}{1+x^2}$$$$\underset{x\to -0}{\lim }{f}^{\prime }\left(x\right)=-\frac{\pi }{2}-0=-\frac{\pi }{2}$$$$\underset{x\to +0}{\lim }{f}^{\prime }\left(x\right)=\frac{\pi }{2}-0=\frac{\pi }{2}$$$$since\underset{x\to -0}{\lim }{f}^{\prime }\left(x\right)\ne \underset{x\to +0}{\lim }{f}^{\prime }\left(x\right)$$$$\Rightarrow f\left(x\right)isnotdifferentiableatx=0.$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com