Discuss the continuity of 'f' in [0, 2], where `f(x) = {{:(|4x - 5|[x],"for",x gt 1),([cos pi x],"for",x le 1):}`, where [x] is greastest integer not greater than x.

Answer

To discuss the continuity of the function \( f(x) \) defined as:

\[ f(x) = \begin{cases} |4x - 5| & \text{for } x > 1 \\ \cos(\pi x) & \text{for } x \leq 1 \end{cases} \]

on the interval \([0, 2]\), we need to check the continuity at the points where the definition of the function changes, specifically at \( x = 1 \). We will also check the continuity at the endpoints \( x = 0 \) and \( x = 2 \).

### Step 1: Check continuity at \( x = 1 \)

To check continuity at \( x = 1 \), we need to find the following:

1. \( f(1) \)
2. \( \lim_{x \to 1^-} f(x) \) (left-hand limit)
3. \( \lim_{x \to 1^+} f(x) \) (right-hand limit)

#### 1. Calculate \( f(1) \)

Since \( x = 1 \) falls under the case \( x \leq 1 \):

\[ f(1) = \cos(\pi \cdot 1) = \cos(\pi) = -1 \]

#### 2. Calculate \( \lim_{x \to 1^-} f(x) \)

For \( x \) approaching 1 from the left (values less than 1), we use the definition \( f(x) = \cos(\pi x) \):

\[ \lim_{x \to 1^-} f(x) = \cos(\pi \cdot 1) = \cos(\pi) = -1 \]

#### 3. Calculate \( \lim_{x \to 1^+} f(x) \)

For \( x \) approaching 1 from the right (values greater than 1), we use the definition \( f(x) = |4x - 5| \):

\[ \lim_{x \to 1^+} f(x) = |4 \cdot 1 - 5| = |4 - 5| = | -1 | = 1 \]

### Step 2: Compare limits and function value at \( x = 1 \)

Now we compare the limits and the function value:

- \( f(1) = -1 \)
- \( \lim_{x \to 1^-} f(x) = -1 \)
- \( \lim_{x \to 1^+} f(x) = 1 \)

Since \( \lim_{x \to 1^-} f(x) \neq \lim_{x \to 1^+} f(x) \), the function is **not continuous at \( x = 1 \)**.

### Step 3: Check continuity at \( x = 0 \)

For \( x = 0 \):

\[ f(0) = \cos(\pi \cdot 0) = \cos(0) = 1 \]

Since there are no other points to check for \( x < 0 \) in our interval, and the function is defined and continuous for all \( x < 1 \), \( f(x) \) is continuous at \( x = 0 \).

### Step 4: Check continuity at \( x = 2 \)

For \( x = 2 \):

\[ f(2) = |4 \cdot 2 - 5| = |8 - 5| = |3| = 3 \]

Since there are no points to check for \( x > 2 \) in our interval, and the function is defined and continuous for all \( x > 1 \), \( f(x) \) is continuous at \( x = 2 \).

### Conclusion

The function \( f(x) \) is continuous at \( x = 0 \) and \( x = 2 \), but it is **not continuous at \( x = 1 \)**.

### Summary of Continuity

- Continuous at \( x = 0 \)
- Not continuous at \( x = 1 \)
- Continuous at \( x = 2 \)