Question

Let \( f(x) = \begin{cases} 3x + 6, & \text{if } x \ge c \\ x^{2} - 3x - 1, & \text{if } x<c \end{cases} \), where \( x \in \mathbb{R} \) and \( c \) is a constant. The values of \( c \) for which \( f \) is continuous on \( \mathbb{R} \) are:

• A. -7, 1
• B. 1, 3
• C. -1, 7
• D. -1, 6
• E. 2, -3

Hint:

Continuity at a boundary \(c\) for functions \(g(x)\) and \(h(x)\) simply means solving the equation \(g(c) = h(c)\). This is effectively finding the x-coordinates where the two graphs intersect.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
For a piecewise function to be continuous everywhere, it must specifically be continuous at the "boundary" point where the definition changes. This means the left-hand limit, right-hand limit, and the function value at \(x=c\) must all be equal.

Step 2: Key Formula or Approach:
Set \(\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x)\).

Step 3: Detailed Explanation:
1. Find the left-hand limit (LHL) as \(x \to c\): \[ \text{LHL} = c^2 - 3c - 1 \]
2. Find the right-hand limit (RHL) and function value \(f(c)\): \[ \text{RHL} = 3c + 6 \]
3. Equate LHL and RHL for continuity: \[ c^2 - 3c - 1 = 3c + 6 \]
4. Rearrange into a quadratic equation: \[ c^2 - 6c - 7 = 0 \]
5. Factor the quadratic: \[ (c - 7)(c + 1) = 0 \] \[ c = 7 \text{ or } c = -1 \]

Step 4: Final Answer
The values of \(c\) for which \(f\) is continuous are -1 and 7.