The correct option is f has local maximum.

The function shown in the image is:
$$f\left(x\right)=\left(\frac{\left|x\right|}{2}-x\right)\times \left(x-\frac{\left|x\right|}{2}\right)$$

We can simplify the expression by factoring out a negative sign from the first term:
$$\frac{\left|x\right|}{2}-x=-\left(x-\frac{\left|x\right|}{2}\right)$$
Substituting this back into the function gives:
$$f\left(x\right)=-{\left(x-\frac{\left|x\right|}{2}\right)}^2$$

Now, let's analyze the behavior of the function by expanding it piecewise based on the definition of the absolute value function $\left|x\right|$:
1. For $x\ge 0$, we have $\left|x\right|=x$:
$$f\left(x\right)=-{\left(x-\frac{x}{2}\right)}^2=-{\left(\frac{x}{2}\right)}^2=-\frac{x^2}{4}$$
2. For $x<0$, we have $\left|x\right|=-x$:
$$f\left(x\right)=-{\left(x-\frac{-x}{2}\right)}^2=-{\left(x+\frac{x}{2}\right)}^2=-{\left(\frac{3x}{2}\right)}^2=-\frac{9x^2}{4}$$

Let's evaluate the function values:
- At $x=0$, we have $f\left(0\right)=0$.
- For any $x>0$, the value is $-\frac{x^2}{4}$, which is strictly negative ($f\left(x\right)<0$).
- For any $x<0$, the value is $-\frac{9x^2}{4}$, which is also strictly negative ($f\left(x\right)<0$).

Since $f\left(x\right)<f\left(0\right)$ for all $x\ne 0$ in the neighborhood of $x=0$, the function reaches a local maximum value of $0$ at $x=0$. Therefore, f has a local maximum.