Question

If a function $f\left(x\right)$ is defined as $f\left(x\right)=\{\begin{array}{c}-x,x<0\\ x^2,0\le x\le 1\\ x^2-x+1,x>1\end{array}$, then-

• A. $f\left(x\right)$ is differentiable at $x=0$ and $x=1$
• B. $f\left(x\right)$ is differentiable at $x=0$ but not at $x=1$
• C. $f\left(x\right)$ is differentiable at $x=1$ but not at $x=0$
• D. $f\left(x\right)$ is not differentiable at $x=0$ and $x=1$

Answer 1

The correct option is D $f\left(x\right)$ is not differentiable at $x=0$ and $x=1$

If computed the values of the function f(x) at points 0 and 1(where the curve changes), we find the function to be continuous and $f\left(0\right)=0$ and $f\left(1\right)=1.$
But, when differentiated, $f^{\prime }\left(0^{-}\right)$ = -1 while $f^{\prime }\left(0^{+}\right)=2\left(0\right)=0$
For $x=1,f^{\prime }\left(1^{-}\right)=2\left(1\right)=2$ while $f^{\prime }\left(1^{+}\right)=2\left(1\right)-1=1$
Hence we find that $f^{\prime }\left(0^{-}\right)\ne f^{\prime }\left(0^{+}\right)$ and similarly, $f^{\prime }\left(1^{-}\right)\ne f^{\prime }\left(1^{+}\right)$
Thus, function is not differential at both these points.