If f(x) = \ cases e^ x , & x ≤ 0 |1 - x|, & x > 0 cases ), then

f\left(x\right) = \left\{

\begin{cases}e^{x}, & x \le 0 \\ \left|1 - x\right|, & x > 0\end{cases}$\right), then $If f(x) = \ cases e^ x , & x ≤ 0 |1 - x|, & x > 0 cases ), then$

Options

A $f$ \left(x\right) $is differentiable at$ x = 0
B $f$ \left(x\right) $is continuous at$ x = 0, 1
C $f$ \left(x\right) $is differentiable at$ x = 1
DNone of the above

Correct answer

B. $f$ \left(x\right) $is continuous at$ x = 0, 1

Step-by-step solution

f\left(x\right) = \left\{ $\begin{cases}e^{x}, & x \le 0 \\ 1 - x, & 0 1\end{cases}$ \right)

$If f(x) = \ cases e^ x , & x ≤ 0 |1 - x|, & x > 0 cases ), then$

$RHD: f^{'}\left(0\right) = \lim_{h \to 0}\dfrac{f\left(0 + h\right) - f\left(0\right)}{h} = \lim_{h \to 0} - \left[\dfrac{h - 1}{h}\right] = - 1$

$LHD: f^{'}\left(0\right) = \lim_{h \to 0}\dfrac{f\left(0 - h\right) - f\left(0\right)}{- h} = \lim_{h \to 0}\dfrac{e^{- h} - 1}{- h} = 1$

So, it is not differentiable at $x = 0$ .

Similarly, it is not differentiable at $x = 1$ but it is continuous at $x = 0$ and $1$ .

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