Question

If the function \( f(x) \), defined below, is continuous on the interval \([0,8]\), then: \[ f(x) = \begin{cases} x^2 + ax + b, & 0 \leq x < 2 \\ 3x + 2, & 2 \leq x \leq 4 \\ 2ax + 5b, & 4 < x \leq 8 \end{cases} \]

• A. a = 3, b = −2
• B. a = −3, b = 2
• C. a = −3, b = −2
• D. a = 3, b = 2

Hint:

Continuity at a point requires matching left-hand and right-hand limits.

Answer 1

The Correct Option is A

Since \( f(x) \) is continuous on \([0,8]\), it must be continuous at \( x = 2 \) and \( x = 4 \). At \( x = 2 \), \[ \lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) \] \[ \lim_{x \to 2^-} (x^2 + ax + b) = \lim_{x \to 2^+} (3x + 2) \] \[ 4 + 2a + b = 3(2) + 2 \] \[ 2a + b = 4 \quad {(Equation 1)} \] At \( x = 4 \), \[ \lim_{x \to 4^-} f(x) = \lim_{x \to 4^+} f(x) \] \[ \lim_{x \to 4^-} (3x + 2) = \lim_{x \to 4^+} (2ax + 5b) \] \[ 3(4) + 2 = 2a(4) + 5b \] \[ 8a + 5b = 14 \quad {(Equation 2)} \] Solving Equations 1 and 2, we get: \[ a = 3, \quad b = -2 \]

Answer 2

Step 1: Understanding the problem
We are given a piecewise function \( f(x) \) that is defined on the interval \([0,8]\). The function is continuous on this interval. We need to find the values of constants \( a \) and \( b \) that make the function continuous at the points where the function changes its form: at \( x = 2 \) and \( x = 4 \).

Step 2: Apply the continuity condition at \( x = 2 \)
At \( x = 2 \), the function changes from \( f(x) = x^2 + ax + b \) for \( 0 \leq x < 2 \) to \( f(x) = 3x + 2 \) for \( 2 \leq x \leq 4 \). For the function to be continuous at \( x = 2 \), the left-hand limit and the right-hand limit must be equal. This means: \[ \lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) \] Substitute \( x = 2 \) into both expressions: - From the first piece, \( f(x) = x^2 + ax + b \), we get: \[ f(2) = 2^2 + 2a + b = 4 + 2a + b \] - From the second piece, \( f(x) = 3x + 2 \), we get: \[ f(2) = 3(2) + 2 = 6 + 2 = 8 \] Equating the two expressions for continuity at \( x = 2 \): \[ 4 + 2a + b = 8 \] Simplify: \[ 2a + b = 4 \quad \text{(Equation 1)} \] Step 3: Apply the continuity condition at \( x = 4 \)
At \( x = 4 \), the function changes from \( f(x) = 3x + 2 \) for \( 2 \leq x \leq 4 \) to \( f(x) = 2ax + 5b \) for \( 4 < x \leq 8 \). For the function to be continuous at \( x = 4 \), the left-hand limit and the right-hand limit must be equal: \[ \lim_{x \to 4^-} f(x) = \lim_{x \to 4^+} f(x) \] Substitute \( x = 4 \) into both expressions: - From the second piece, \( f(x) = 3x + 2 \), we get: \[ f(4) = 3(4) + 2 = 12 + 2 = 14 \] - From the third piece, \( f(x) = 2ax + 5b \), we get: \[ f(4) = 2a(4) + 5b = 8a + 5b \] Equating the two expressions for continuity at \( x = 4 \): \[ 8a + 5b = 14 \quad \text{(Equation 2)} \] Step 4: Solve the system of equations
We now have the system of two equations: \[ 2a + b = 4 \quad \text{(Equation 1)} \] \[ 8a + 5b = 14 \quad \text{(Equation 2)} \] Solve Equation 1 for \( b \): \[ b = 4 - 2a \] Substitute this into Equation 2: \[ 8a + 5(4 - 2a) = 14 \] Simplify: \[ 8a + 20 - 10a = 14 \] \[ -2a + 20 = 14 \] \[ -2a = -6 \quad \Rightarrow \quad a = 3 \] Now substitute \( a = 3 \) into Equation 1 to find \( b \): \[ 2(3) + b = 4 \quad \Rightarrow \quad 6 + b = 4 \quad \Rightarrow \quad b = -2 \] Step 5: Final Answer
The values of \( a \) and \( b \) are:

a = 3, b = −2