Question

The function $f(x)=\{\begin{array}{ll}{x}^2& \text{for }x<1\\ 2-x& \text{for }x\ge 1\end{array}$ is

• A. not differentiable at $x=1$
• B. differentiable at $x=1$
• C. not continuous at $x=1$
• D. none of these

Answer 1

Answer: (A)

$f(x)=\{\begin{array}{ll}{x}^2& \text{for }x<1\\ 2-x& \text{for }x\ge 1\end{array}$
$\underset{x\to 1^{-}}{\lim }f\left(x\right)=\underset{x\to 1^{+}}{\lim }f\left(x\right)=f\left(1\right)$
$\therefore \underset{x\to 1^{-}}{\lim }{x}^2=\underset{x\to 1^{+}}{\lim }2-x=2-1$
Hence, $f(x)$ is continuous at $x=1$
Now, $f^{\prime }(x)=\{\begin{array}{ll}2x& \text{for }x<1\\ -1& \text{for }x\ge 1\end{array}$
$\therefore \underset{x\to 1^{-}}{\lim }{f}^{\prime }\left(x\right)=\underset{x\to 1^{-}}{\lim }2x=2$
$\underset{x\to 1^{+}}{\lim }{f}^{\prime }\left(x\right)=\underset{x\to 1^{-}}{\lim }2x=2\underset{x\to 1^{+}}{\lim }{f}^{\prime }\left(x\right)=\underset{x\to 1^{+}}{\lim }\left(-x\right)=-1$
$\Rightarrow \underset{x\to 1^{-}}{\lim }f\left(x\right)\ne \underset{x\to 1^{+}}{\lim }{f}^{\prime }\left(x\right)$
$i.e.,L.H.D\ne R.H.D.$
Hence. $f(x)$ is.no] differetiable at $x=1$