VerifiedHint:
A function is continuous at a point $x = c$ if:
1. $f(c)$ is defined.
2. $lim_{{x \to c^-}} f(x)$ (left-hand limit) exists.
3. $lim_{{x \to c^+}} f(x)$ (right-hand limit) exists.
4. $lim_{{x \to c^-}} f(x) = \lim_{{x \to c^+}} f(x) = f(c)$.
Step-by-step Solution:
1. At $x = 0$:
$f(0)$ is clearly defined as 0.
Left-hand limit: $lim_{{x \to 0^-}} f(x) = 0$
Right-hand limit: $lim_{{x \to 0^+}} f(x) = 0$
Since both the limits are equal to $f(0)$, the function is continuous at $x = 0$.
2. At $x = 1$:
$f(1)$ is defined as 1.
Left-hand limit: $lim_{{x \to 1^-}} f(x) = 1$
Right-hand limit: $lim_{{x \to 1^+}} f(x) = 5$
The left-hand limit is not equal to the right-hand limit, so the function is discontinuous at $x = 1$.
3. At $x = 2$:
$f(2)$ is defined as 5.
Left-hand limit: $lim_{{x \to 2^-}} f(x) = 5$
Right-hand limit: $lim_{{x \to 2^+}} f(x) = 5$
Both the limits are equal to $f(2)$, so the function is continuous at $x = 2$.
Points of Discontinuity:
A function is discontinuous at a point if it is not continuous at that point. Therefore, the only point of discontinuity of $f$ is $x = 1$.
Note:
The function $f(x)$ is continuous everywhere except when it changes its definition rule, which is $x = 1$. Whenever you encounter piecewise-defined functions, always check the points where the function definition changes to find possible points of discontinuity.
