Satisfying Conditions for Differentiable Function at \( x = -2 \)

Let's solve for \( a \) and \( b \) so that the function \[ f(x) = \begin{cases} ax-4 & \text{if } x \leq -2 \\ 2bx^2 + x - 8 & \text{if } x > -2 \\ \end{cases} \] is **differentiable** at \( x=-2 \). Differentiability at a point requires two things: 1. **Continuity** at \( x=-2 \): \( f(-2^-) = f(-2^+) \) 2. **Equality of derivatives** at \( x=-2 \): \( f'_1(-2) = f'_2(-2) \) Let's check both conditions step by step. --- ### 1. Continuity at \( x = -2 \) Set \( f(-2^-) = f(-2^+) \): - For \( x \leq -2 \): \( f(x) = ax - 4 \) \( f(-2^-) = a(-2) - 4 = -2a - 4 \) - For \( x > -2 \): \( f(x) = 2b x^2 + x - 8 \) \( f(-2^+) = 2b(-2)^2 + (-2) - 8 = 8b - 2 - 8 = 8b - 10 \) Equate them: \[ -2a - 4 = 8b - 10 \] \[ -2a = 8b - 6 \] \[ a = -4b + 3 \] --- ### 2. Equality of the derivatives at \( x = -2 \) First, find the derivatives for both branches: - \( f_1(x) = ax - 4 \implies f_1'(x) = a \) - \( f_2(x) = 2b x^2 + x - 8 \implies f_2'(x) = 4b x + 1 \) At \( x=-2 \): - Left derivative: \( f_1'(-2) = a \) - Right derivative: \( f_2'(-2) = 4b(-2) + 1 = -8b + 1 \) Set them equal: \[ a = -8b + 1 \] --- ### 3. Solve the system You have two equations: 1. \( a = -4b + 3 \) 2. \( a = -8b + 1 \) Set equal: \[ -4b + 3 = -8b + 1 \] \[ -4b + 3 + 8b - 1 = 0 \] \[ 4b + 2 = 0 \] \[ 4b = -2 \implies b = -\frac{1}{2} \] Now, substitute back to get \( a \): \[ a = -4b + 3 = -4 \left(-\frac{1}{2}\right) + 3 = 2 + 3 = 5 \] --- ### **Final Answer** \[ \boxed{a = 5, \quad b = -\frac{1}{2}} \] will make \( f(x) \) differentiable at \( x = -2 \).