The function `f(x)= {(5x-4 ", " 0 lt x le 1 ),( 4x^3-3x", " 1 lt x lt 2):}`

Options

A. continuous at x=1
B. discontinuous at x= 1
C. continuous at x= 2
D. None of these

Answer

To determine the continuity of the function \( f(x) \) at the points \( x = 1 \) and \( x = 2 \), we will follow these steps:

### Step 1: Define the function
The function is defined as:
\[ f(x) = \begin{cases} 5x - 4 & \text{for } 0 < x \leq 1 \\ 4x^3 - 3x & \text{for } 1 < x < 2 \end{cases} \]

### Step 2: Check continuity at \( x = 1 \)
To check continuity at \( x = 1 \), we need to evaluate the left-hand limit, right-hand limit, and the function value at \( x = 1 \).

#### Step 2.1: Calculate the left-hand limit as \( x \) approaches 1
\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (5x - 4) \]
Substituting \( x = 1 \):
\[ = 5(1) - 4 = 5 - 4 = 1 \]

#### Step 2.2: Calculate the right-hand limit as \( x \) approaches 1
\[ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (4x^3 - 3x) \]
Substituting \( x = 1 \):
\[ = 4(1)^3 - 3(1) = 4 - 3 = 1 \]

#### Step 2.3: Calculate the function value at \( x = 1 \)
Since \( f(x) \) is defined for \( 0 < x \leq 1 \):
\[ f(1) = 5(1) - 4 = 1 \]

### Step 3: Compare the limits and function value
We have:
\[ \lim_{x \to 1^-} f(x) = 1, \quad \lim_{x \to 1^+} f(x) = 1, \quad f(1) = 1 \]
Since both limits are equal and equal to the function value, we conclude that:
\[ f(x) \text{ is continuous at } x = 1 \]

### Step 4: Check continuity at \( x = 2 \)
To check continuity at \( x = 2 \), we need to evaluate the left-hand limit and the function value at \( x = 2 \).

#### Step 4.1: Calculate the left-hand limit as \( x \) approaches 2
\[ \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (4x^3 - 3x) \]
Substituting \( x = 2 \):
\[ = 4(2)^3 - 3(2) = 4(8) - 6 = 32 - 6 = 26 \]

#### Step 4.2: Calculate the function value at \( x = 2 \)
The function \( f(x) \) is not defined at \( x = 2 \) since it is only defined for \( 1 < x < 2 \).

### Step 5: Conclusion for continuity at \( x = 2 \)
Since the left-hand limit as \( x \) approaches 2 is 26 and the function is not defined at \( x = 2 \), we conclude that:
\[ f(x) \text{ is discontinuous at } x = 2 \]

### Final Answer
- The function \( f(x) \) is continuous at \( x = 1 \).
- The function \( f(x) \) is discontinuous at \( x = 2 \).

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