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 \( f(x) \) defined as:

\[ f(x) = \begin{cases} 5x - 4 & \text{for } 0 < x \leq 1 \\ 4x^3 - 3x & \text{for } 1 < x < 2 \end{cases} \]

we need to check the continuity at the point \( x = 1 \).

### Step 1: Find the left-hand limit as \( x \) approaches 1.

The left-hand limit is calculated using the expression for \( f(x) \) when \( x \) is approaching 1 from the left side (i.e., \( x \to 1^- \)):

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

Substituting \( x = 1 \):

\[ = 5(1) - 4 = 5 - 4 = 1 \]

### Step 2: Find the right-hand limit as \( x \) approaches 1.

The right-hand limit is calculated using the expression for \( f(x) \) when \( x \) is approaching 1 from the right side (i.e., \( x \to 1^+ \)):

\[ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (4x^3 - 3x) \]

Substituting \( x = 1 \):

\[ = 4(1)^3 - 3(1) = 4 - 3 = 1 \]

### Step 3: Evaluate the function at \( x = 1 \).

Now, we need to find the value of the function at \( x = 1 \):

\[ f(1) = 5(1) - 4 = 1 \]

### Step 4: Check for continuity at \( x = 1 \).

For the function to be continuous at \( x = 1 \), the following condition must be satisfied:

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

From our calculations:

- Left-hand limit: \( \lim_{x \to 1^-} f(x) = 1 \)
- Right-hand limit: \( \lim_{x \to 1^+} f(x) = 1 \)
- Function value: \( f(1) = 1 \)

Since all three values are equal, we conclude that the function is continuous at \( x = 1 \).

### Conclusion

Thus, the function \( f(x) \) is continuous at \( x = 1 \).