Discuss the continutily of the following fuction at x=1 and 2
`f(x)={(5x-4 ", " x le1),(4x^2-3x", " 1 ltx lt2),(3x+4", "x ge 2):}`

Answer

To discuss the continuity of the given piecewise function at \( x = 1 \) and \( x = 2 \), we will follow these steps:

### Step 1: Define the function
The function is defined as follows:
\[ f(x) = \begin{cases} 5x - 4 & \text{if } x \leq 1 \\ 4x^2 - 3x & \text{if } 1 < x < 2 \\ 3x + 4 & \text{if } x \geq 2 \end{cases} \]

### Step 2: Check continuity at \( x = 1 \)
To check the continuity at \( x = 1 \), we need to verify three conditions:
1. \( f(1) \) is defined.
2. \( \lim_{x \to 1^-} f(x) \) exists.
3. \( \lim_{x \to 1^+} f(x) \) exists.

**Calculating \( f(1) \)**:
Since \( x = 1 \) falls in the first case of the piecewise function:
\[ f(1) = 5(1) - 4 = 1 \]

**Calculating \( \lim_{x \to 1^-} f(x) \)**:
As \( x \) approaches 1 from the left:
\[ \lim_{x \to 1^-} f(x) = 5(1) - 4 = 1 \]

**Calculating \( \lim_{x \to 1^+} f(x) \)**:
As \( x \) approaches 1 from the right, we use the second piece of the function:
\[ \lim_{x \to 1^+} f(x) = 4(1)^2 - 3(1) = 4 - 3 = 1 \]

Since all three values are equal:
\[ f(1) = \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = 1 \]
Thus, \( f(x) \) is continuous at \( x = 1 \).

### Step 3: Check continuity at \( x = 2 \)
Now, we will check the continuity at \( x = 2 \) using the same three conditions.

**Calculating \( f(2) \)**:
Since \( x = 2 \) falls in the third case of the piecewise function:
\[ f(2) = 3(2) + 4 = 6 + 4 = 10 \]

**Calculating \( \lim_{x \to 2^-} f(x) \)**:
As \( x \) approaches 2 from the left, we use the second piece of the function:
\[ \lim_{x \to 2^-} f(x) = 4(2)^2 - 3(2) = 16 - 6 = 10 \]

**Calculating \( \lim_{x \to 2^+} f(x) \)**:
As \( x \) approaches 2 from the right:
\[ \lim_{x \to 2^+} f(x) = 3(2) + 4 = 6 + 4 = 10 \]

Since all three values are equal:
\[ f(2) = \lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = 10 \]
Thus, \( f(x) \) is continuous at \( x = 2 \).

### Conclusion
The function \( f(x) \) is continuous at both \( x = 1 \) and \( x = 2 \).

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