Question

Investigate the following function from the point of view of its differentiability. Does.the differential coefficient of the function exist at x=1? $f\left(x\right)=\{\begin{array}{cc}x,& x<1\\ 2-x,& 1\le x\le 2\\ -2+3x-x^2,& x>2\end{array}$

Answer 1

We test the function $f\left(x\right)$ for differentiability at $x=0$ and $x=$ Only.
For other values of $x,f\left(x\right)$ can check that $f\left(x\right)$ is continuous as $x=0$ and $x=1$.
We now first test the differentiability at x=0.$L{f}^{\prime }\left(0\right)=\underset{h\to 0}{lim}\frac{f(0-h)-f\left(0\right)}{-h}=\underset{h\to 0}{lim}\frac{-(0+h)-0}{-h}=-1.$
$R{f}^{\prime }\left(0\right)=\underset{h\to 0}{lim}\frac{f(0+h)-f\left(0\right)}{h}=\underset{h\to 0}{lim}\frac{{(0+h)}^2-0}{h}=\underset{h\to 0}{lim}h=0.$
Since $L{f}^{\prime }\left(0\right)\ne R{f}^{\prime }\left(0\right),$ the function is not difterentiable at $x=0$.
Again $L{f}^{\prime }\left(1\right)=\underset{h\to 0}{lim}\frac{f(1-h)-f\left(1\right)}{-h}$ $=\underset{h\to 0}{lim}\frac{{(1-h)}^2-1}{-h}=\underset{h\to 0}{lim}\frac{-2h+h^2}{-h}=\underset{h\to 0}{lim}(2-h)=2$
and $R{f}^{\prime }\left(1\right)=\underset{h\to 0}{lim}\frac{f{(1+h)}^3-(1+h)+1-1}{h}=\underset{h\to 0}{lim}\frac{2h+3h^2+h^3}{h}=\underset{h\to 0}{lim}(2+3h+h^3)=2.$
Hence $f^{\prime }\left(1\right)$ exists i.e.,function is differentiable at $x=1.$