Question

Discuss the differentiability of the function $f\left(x\right)=\{\begin{array}{cc}2x-1,& x<\frac{1}{2}\\ 3-6x,& x\ne \frac{1}{2}\end{array}$ at $x=\frac{1}{2}.$

OR For what value of k, is $f\left(x\right)=\{\begin{array}{cc}\frac{\sqrt{3}sinx+cosx}{x+\frac{\pi }{6}},& x\ne -\frac{\pi }{6}\\ k,x=-\frac{\pi }{6}& \end{array}$ continuous at $x=-\frac{\pi }{6}?$

Answer 1

Note that $f\left(1/2\right)=3-6\times \frac{1}{2}=0$

Now Lf'$\left(1/2\right)=\stackrel{lim}{h\to 0^{-}}\frac{f(\frac{1}{2}-h)-f\left(\frac{1}{2}\right)}{-h}=\stackrel{lim}{h\to 0^{-}}\frac{[2(\frac{1}{2}-h)-1]-0}{-h}=2$

And, Rf' $\left(1/2\right)=\stackrel{lim}{h\to 0^{+}}\frac{f(\frac{1}{2}+h)-f\left(\frac{1}{2}\right)}{h}=\stackrel{lim}{h\to 0^{-}}\frac{[3-6(\frac{1}{2}+h)]-0}{h}=-6$

Since Rf' $\left(1/2\right)=L{f}^{\prime }\left(1/2\right)$ so, f isn't differentiable at $x=1/2.$

OR Here $f(-\frac{\pi }{6})=k.$

Also, $\stackrel{lim}{x\to -\frac{\pi }{6}}\frac{\sqrt{3}sinx+cosx}{x+\frac{\pi }{6}}=\stackrel{lim}{x+\frac{\pi }{6}\to 0}\frac{2sin(x+\frac{\pi }{6})}{x+\frac{\pi }{6}}=2\times 1=2$

For the continuity of $f\left(x\right)$ at $x=-\frac{\pi }{6},$ we have $f(-\frac{\pi }{6})=\stackrel{lim}{x+\frac{\pi }{6}\to 0}f\left(x\right)\therefore k=2.$