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:
A piecewise function is made up of different sub-functions over different intervals.
The sub-functions \( 3x + 6 \) (a line) and \( x^2 - 3x - 1 \) (a parabola) are polynomials, which are continuous everywhere on their own domains.
The only possible point of discontinuity for the entire function \( f(x) \) is at the boundary point \( x = c \), where the rule defining the function changes.
To ensure \( f(x) \) is continuous on the entire real line \( \mathbb{R} \), we must enforce continuity specifically at this boundary \( x = c \).
Step 2: Key Formula or Approach:
For a function \( f(x) \) to be continuous at a point \( x = c \), the following three conditions must hold:
1. \( f(c) \) is defined.
2. The limit as \( x \) approaches \( c \) exists, meaning the left-hand limit equals the right-hand limit: \( \lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) \).
3. The limit equals the function value: \( \lim_{x \to c} f(x) = f(c) \).
Step 3: Detailed Explanation:
Let's evaluate the left-hand limit, right-hand limit, and the exact function value at \( x = c \).
1. Left-Hand Limit (LHL): Approaching \( c \) from the left (\( x<c \)), we use the bottom piece:
\[ \lim_{x \to c^-} f(x) = \lim_{x \to c^-} (x^2 - 3x - 1) = c^2 - 3c - 1 \] 2. Right-Hand Limit (RHL): Approaching \( c \) from the right (\( x>c \)), we use the top piece:
\[ \lim_{x \to c^+} f(x) = \lim_{x \to c^+} (3x + 6) = 3c + 6 \] 3. Function Value at \( c \): Since the top condition is \( x \ge c \), we plug \( x = c \) into the top piece:
\[ f(c) = 3(c) + 6 = 3c + 6 \] For the function to be continuous at \( x = c \), the LHL must exactly equal the RHL (and thus equal \( f(c) \)):
\[ \lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) \] \[ c^2 - 3c - 1 = 3c + 6 \] We must solve this equation for \( c \). Move all terms to one side to form a standard quadratic equation equal to zero:
\[ c^2 - 3c - 3c - 1 - 6 = 0 \] \[ c^2 - 6c - 7 = 0 \] Now, factor the quadratic equation. We look for two factors of -7 that add up to -6. These are -7 and +1:
\[ (c - 7)(c + 1) = 0 \] Set each factor to zero to find the possible values for \( c \):
\[ c - 7 = 0 \implies c = 7 \] \[ c + 1 = 0 \implies c = -1 \] Therefore, the function will be continuous everywhere if the constant \( c \) is either -1 or 7.
Step 4: Final Answer:
The values of \( c \) are -1, 7.