5. continuity and differentiation — ગણિત ધોરણ 12 સાયન્સ — Question

$ = \left\{ {\begin{array}{*{20}{c}}{ - 2x + 1,}&{x < 0}&{}\\{x - x + 1,}&{0 \le x < 1}& = \\{x + x - 1,}&{x \ge 1}&{}\end{array}} \right.\left\{ {\begin{array}{*{20}{c}}{ - 2x + 1,}&{x < 0}\\1&{0 \le x < 1}\\{2x - 1,}&{x \ge 1}\end{array}} \right.$

Clearly, $\mathop {\lim }\limits_{x \to {0^ - }} f(x) = 1,\,\,\mathop {\lim }\limits_{x \to {0^ + }} f(x) = 1,\,\,\mathop {\lim }\limits_{x \to {1^ - }} f(x) = 1$

and $\mathop {\lim }\limits_{x \to {1^ + }} f(x) = 1$.

So, $f(x)$ is continuous at $x = 0,\,\,1.$

Now $f'(x) = \left\{ {\begin{array}{*{20}{l}}{ - 2,\,\,\,\,x < 0}\\{\,\,0,\,\,\,\,\,0 \le x < 1}\\{\,\,2,\,\,\,\,\,x \ge 1}\end{array}} \right.$

Here $x = 0$, $f'({0^ + }) = 0$ while $f'({0^ - }) = - 2$

and at $x = 1$, $f'({1^ + }) = 2$ while $f'({1^ - }) = 0$

Thus, $f(x)$ is not differentiable at $x = 0$ and $1.$