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

First, find the critical points of $|x|^2 - 5|x|$.

Let $y = |x|^2 - 5|x|$. Then, $y = x^2 - 5x$ when $x \geq 0$, and $y = x^2 + 5x$ when $x < 0$.

Finding the derivative of $y$ with respect to $x$:

For $x \geq 0$:

$y' = \frac{d}{dx} (x^2 - 5x) = 2x - 5$

For $x < 0$:

$y' = \frac{d}{dx} (x^2 + 5x) = 2x + 5$

Setting these derivatives equal to zero to find critical points:

For $x \geq 0$:

$2x - 5 = 0$

$2x = 5$

$x = \frac{5}{2}$

For $x < 0$:

$2x + 5 = 0$

$2x = -5$

$x = -\frac{5}{2}$

So, the critical points are $x = \frac{5}{2}$ and $x = -\frac{5}{2}$.

Now, evaluate $f(x)$ at these critical points:

For $x = \frac{5}{2}$:

$f\left(\frac{5}{2}\right) = \min\left(|\frac{5}{2}|^{2} - 5|\frac{5}{2}|, 1\right)$

$= \min\left(\frac{25}{4} - \frac{25}{2}, 1\right)$

$= \min\left(-\frac{25}{4}, 1\right)$

$= -\frac{25}{4}$

For $x = -\frac{5}{2}$:

$f\left(-\frac{5}{2}\right) = \min\left(|-\frac{5}{2}|^{2} - 5|-\frac{5}{2}|, 1\right)$

$= \min\left(\frac{25}{4} + \frac{25}{2}, 1\right)$

$= \min\left(\frac{75}{4}, 1\right)$

$= 1$

So, $f(x)$ is non-differentiable at $x = \frac{5}{2}$ because $f(x)$ has a sharp turn at that point, and it's also non-differentiable at $x = -\frac{5}{2}$ because the function has a sharp corner there.

Therefore, $\lambda = 2$, and $\lambda + 13 = 2 + 13 = 15$.