Question

If $f\left(x\right)=\text{min}\left\{x^2,-x+1,sgn|-x|\right\}$ then $f\left(x\right)$ is
(where $\text{sgn}\left(x\right)$ denotes signum function of $x$)

• A. continuous at $x=0$
• B. discontinuous at one point
• C. non-differentiable at $3$ points
• D. differentiable at $x=\frac{1}{2}$

Answer 1

Answer: (A), (D)

differentiable at $x=\frac{1}{2}$

By the figure given above we can say that $f\left(x\right)$ graph will be

$\text{The curves}{x}^2\text{and}-x+1\text{intersect at}\Rightarrow x^2=-x+1\Rightarrow x^2+x-1=0$
$\Rightarrow x=\frac{1\pm \sqrt{5}}{2}$
$f\left(x\right)$ is continuous for all real numbers but not differentiable at $x=-1,\frac{1+\sqrt{5}}{2}$
So, $f\left(x\right)$ is continous at $x=0$ and differentiable at $x=\frac{1}{2}.$