Question

Let \(f(x) =   \begin{cases}     x+2       & for\ x\lt\ 1 \\     4x-1  & for\ 1\leq x \leq3 \\ x^2+5 & for\ x\gt3   \end{cases}\). Then

• A. f(x) is not continuous at x=-1
• B. f(x) is continuous at x=1
• C. f(x) is continuous at x=3
• D. f(x) is not continuous at x=5
• E. f(x) is not continuous at x=2

Answer 1

The Correct Option is B

For \( f(x) \) to be continuous at a point \( c \), the following must hold:  
1. \( f(c) \) is defined.  
2. \( \lim_{x \to c} f(x) \) exists.  
3. \( \lim_{x \to c} f(x) = f(c) \).  

Let's check the continuity at \( x = 1 \):
- For \( x < 1 \), \( f(x) = x + 2 \). At \( x = 1 \), \( f(x) = 1 + 2 = 3 \).  
- For \( 1 \leq x \leq 3 \), \( f(x) = 4x - 1 \). At \( x = 1 \), \( f(x) = 4(1) - 1 = 3 \).  
Thus, \( f(x) \) is continuous at \( x = 1 \) because the left-hand limit and the function value are equal. \\
 

The correct option is (B) : f(x) is continuous at \(x = 1\)

Answer 2

We have the piecewise function:

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

To check for continuity, we need to examine the points where the function definition changes, i.e., x = 1 and x = 3.

Continuity at x = 1:

Left-hand limit: \( \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x+2) = 1 + 2 = 3 \)

Right-hand limit: \( \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (4x-1) = 4(1) - 1 = 3 \)

Function value at x = 1: \( f(1) = 4(1) - 1 = 3 \)

Since the left-hand limit, right-hand limit, and function value are all equal at x = 1, the function is continuous at x = 1.

Continuity at x = 3:

Left-hand limit: \( \lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} (4x-1) = 4(3) - 1 = 11 \)

Right-hand limit: \( \lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} (x^2+5) = 3^2 + 5 = 9 + 5 = 14 \)

Function value at x = 3: \( f(3) = 4(3) - 1 = 11 \)

Since the left-hand limit and the function value are equal to 11, but the right-hand limit is equal to 14 at x=3, the function is not continuous at x = 3.

Since the options state:

• f(x) is not continuous at x=-1 is incorrect as f(x) is a linear function and continuous from (-infinity, 1)
• f(x) is not continuous at x=5 is incorrect as f(x) is a continuous function for x>3
• f(x) is not continuous at x=2 is incorrect as f(x) is continuous from [1,3]

The function is continuous at x = 1, as the left-hand limit, right-hand limit, and function value are all equal to 3.

Therefore, the answer is f(x) is continuous at x=1.