5. Continuous functions 5.1. Find the points of discontinuity if any for the function \[ f(x)=\left\{\begin{array}{ll}x+1 & \text { if } \ 2 x \geq 2 \ 2 x-1 & \text { if } \ x-1

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To find the points of discontinuity for the piecewise function $$ f(x) $$, we need to analyze the function at the boundaries of the defined intervals. The function is defined as follows:

$$
f(x) = 
\begin{cases} 
x + 1 & \text{if } x \leq 1 \ 
2x - 1 & \text{if } 1 < x \leq 2 \ 
2x & \text{if } x > 2 
\end{cases}
$$

We will check the points $$ x = 1 $$ and $$ x = 2 $$ for continuity.
To check continuity at $$ x = 1 $$, we need to evaluate the left-hand limit, right-hand limit, and the function value at that point:

1. **Left-hand limit** as $$ x $$ approaches 1:
   $$
   \lim_{x \to 1^-} f(x) = 1 + 1 = 2
   $$

2. **Right-hand limit** as $$ x $$ approaches 1:
   $$
   \lim_{x \to 1^+} f(x) = 2(1) - 1 = 1
   $$

3. **Function value** at $$ x = 1 $$:
   $$
   f(1) = 1 + 1 = 2
   $$

Since the left-hand limit (2) does not equal the right-hand limit (1), there is a discontinuity at $$ x = 1 $$.
Next, we check continuity at $$ x = 2 $$:

1. **Left-hand limit** as $$ x $$ approaches 2:
   $$
   \lim_{x \to 2^-} f(x) = 2(2) - 1 = 3
   $$

2. **Right-hand limit** as $$ x $$ approaches 2:
   $$
   \lim_{x \to 2^+} f(x) = 2(2) = 4
   $$

3. **Function value** at $$ x = 2 $$:
   $$
   f(2) = 2(2) - 1 = 3
   $$

Since the left-hand limit (3) does not equal the right-hand limit (4), there is also a discontinuity at $$ x = 2 $$.
The points of discontinuity for the function $$ f(x) $$ are at $$ x = 1 $$ and $$ x = 2 $$.

Continuous functions
5.1. Find the points of discontinuity if any for the function

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