Question:

Let $f(x) = \text{tgn}(x) \cdot \sin(x) \cdot |x|$. The function $\text{tgn}(x)$ is defined as $\text{tgn}(x) = \frac{|x|}{x}$ for $x \neq 0$ and $\text{tgn}(0) = 0$. This is equivalent to $\text{tgn}(x) = 1$ for $x > 0$, $\text{tgn}(x) = -1$ for $x < 0$, and $\text{tgn}(0) = 0$. Thus, $f(x)$ can be written as: $$ f(x) = \begin{cases} \sin(x) \cdot x & \text{if } x > 0 \\ 0 & \text{if } x = 0 \\ -\sin(x) \cdot (-x) & \text{if } x < 0 \end{cases} $$ This simplifies to: $$ f(x) = \begin{cases} x \sin(x) & \text{if } x \ge 0 \\ x \sin(x) & \text{if } x < 0 \end{cases} $$ Therefore, $f(x) = x \sin(x)$ for all $x \in \mathbb{R}$. The function $g(x) = x \sin(x)$ is a product of two continuous functions, $h_1(x) = x$ and $h_2(x) = \sin(x)$. The product of continuous functions is continuous. Thus, $f(x) = x \sin(x)$ is continuous everywhere on $\mathbb{R}$. The number of points in the interval $[0,10]$ where $f(x)$ is discontinuous is 0.