Question

If the derivative of the function \( f(x) = \begin{cases} b x^2 + ax + 4; & x \geq -1 \\ a x^2 + b; & x < -1 \end{cases} \) is everywhere continuous, then
 

• A. \( a = 2, b = 3 \)
• B. \( a = 3, b = 2 \)
• C. \( a = -2, b = -3 \)
• D. \( a = -3, b = -2 \)

Hint:

For piecewise functions, ensure the derivatives from both sides of the function match at the point of interest to ensure continuity.

Answer 1

The Correct Option is A

For the derivative to be continuous, the left-hand and right-hand derivatives must match at \( x = -1 \). This means we need to calculate the derivative of the function on both sides and ensure they match at \( x = -1 \). First, calculate the derivatives of each part of the function: 1. For \( x \geq -1 \), the derivative is: \[ f'(x) = 2b x + a \] 2. For \( x < -1 \), the derivative is: \[ f'(x) = 2a x \] At \( x = -1 \), the derivatives must match: \[ 2b (-1) + a = 2a (-1) \] Simplifying: \[ -a + a = -2a \] Thus, \( a = 2 \). Substituting \( a = 2 \) into the equation: \[ -2b + 2 = -2a \] Solving for \( b \): \[ -2b + 2 = -4 \quad \Rightarrow \quad -2b = -6 \quad \Rightarrow \quad b = 3 \] Thus, the values of \( a \) and \( b \) are \( a = 2 \) and \( b = 3 \).