1. **State the problem:** We have a piecewise function $$f(x) = \begin{cases} 2x + a, & x \leq 3 \\ bx + 4, & x > 3 \end{cases}$$ We need to find values of $a$ and $b$ such that $f(x)$ is differentiable everywhere, especially at $x=3$. 2. **Recall the conditions for differentiability at $x=3$:** - The function must be continuous at $x=3$: $$f(3)^- = f(3)^+$$ - The derivatives from the left and right at $x=3$ must be equal: $$f'(3)^- = f'(3)^+$$ 3. **Apply continuity condition:** From the left side: $$f(3)^- = 2(3) + a = 6 + a$$ From the right side: $$f(3)^+ = b(3) + 4 = 3b + 4$$ Set equal for continuity: $$6 + a = 3b + 4$$ 4. **Apply differentiability condition:** Derivatives: - Left derivative: $f'(x) = 2$ - Right derivative: $f'(x) = b$ Set equal at $x=3$: $$2 = b$$ 5. **Solve for $a$ and $b$:** From differentiability: $$b = 2$$ Substitute into continuity: $$6 + a = 3(2) + 4 = 6 + 4 = 10$$ So, $$a = 10 - 6 = 4$$ **Final answer:** $$a = 4, \quad b = 2$$