Practicing Success
If $f(x)= \begin{cases}\frac{x}{1+|x|}, & |x| \geq 1 \\ \frac{x}{1-|x|}, & |x|<1\end{cases}$ then f(x) is |
discontinuous and non-differentiable at x = -1, 1, 0 discontinuous and non-differentiable at x = -1, whereas continuous and differentiable at x = 0, 1 discontinuous and non-differentiable at x = -1, 1, whereas continuous and differentiable at x = 0 none of these |
discontinuous and non-differentiable at x = -1, 1, whereas continuous and differentiable at x = 0 |
We have, $f(x)= \begin{cases}\frac{x}{1+|x|}, & |x| \geq 1 \\ \frac{x}{1-|x|}, & |x|<1\end{cases}$ $\Rightarrow f(x)= \begin{cases}\frac{x}{1-x}, & x \leq-1 \\ \frac{x}{1+x}, & -1<x<0 \\ \frac{x}{1-x}, & 0<x<1 \\ \frac{x}{1+x}, & x \geq 1\end{cases}$ Clearly, f(x) is discontinuous and hence non-differentiable at x = ±1. It can be easily seen that f(x) is continuous and differentiable at x = 0. |