Question

If $f\left(x\right)=min(|x{|}^2-5|x|,1)$ then $f\left(x\right)$ is non differentiable at $\lambda$ points, then $\lambda +13$ equals

Answer 1

$y=$ min $({\left|x\right|}^2+5\left|x\right|,1)$

Plot the functions

$f\left(x\right)={\left|x\right|}^2+5\left|x\right|$ and $g\left(x\right)=1$

on the same graph paper, and then decide the minimum.

So the graph of min $({\left|x\right|}^2+5\left|x\right|,1)$ is

The function will be non differentiable for all those ${}^{\prime }{x}^{\prime }$ at which the graph takes a sharp turn. From the above graph we can see there are three sharp points (encircled).

Hence $\lambda =3$

$\therefore$ $\lambda +13=3+13=16$