To do this, the absolute value function just returns nonnegative numbers unharmed, and makes negative numbers turn positive. Algebraically, this is accomplished through a piecewise defined function.

$$|x| = \begin {cases} -x &\text {if $x<0$,}\\ x & \text {if $x\ge 0$}. \end {cases}$$

Technically speaking, $|-3| = -(-3)$ and then $-(-3) = 3$. The absolute value function negates negative numbers.

Note: Just because you see a negaitve sign doesn’t mean you have a negative number.

Negating is arithmetic. “Make positive” is not arithmetic.

Graph of $y = A(t) = |t|$.

The absolute value function is a continuous function.

• On $(-\infty , 0)$ we have $| x | = -x$, which is a linear function, which is continuous.
• On $(0, \infty )$ we have $| x | = x$, which is a linear function, which is continuous.

So, the only questions is $0$ itself. The graph certainly suggests continuity, but we want a rigorous explanation.

$| 0 | = 0$, so the only other possibility for $0$ is that it might be a discontinuity. We need to show it is not a discontinuity. We need to show that when $x$ is close to $0$, then also $| x |$ is close to $0$.

Select any small interval you want surrounding the function value $| 0 | = 0$. Like, $(0 - \epsilon , 0 + \epsilon ) = (-\epsilon , \epsilon )$.

Is it possible to find a small inteval around the domain number $0$, like, $(-\delta , \delta )$, so that

$$| x | \in (-\epsilon , \epsilon ) \, \text { whenever } \, x \in (-\delta , \delta )$$

Yes. Just pick $\delta = \epsilon$.

So, $0$ is not a discontinuity and the absolute value function is a continuous function.

Absolute value is a continuous function, which is decreasing on $(-\infty , 0)$ and increasing on $(0, \infty )$. That makes $| 0 | = 0$ the global minimum.

Since the absolute value function is a increasing linear function on $(0, \infty )$, we know that

$$\lim \limits _{x \to \infty }| x | = \infty$$

Therefore, the range is $[0, \infty )$.