1. **State the problem:** We need to find values of $a$ and $b$ such that the function $$g(x) = \begin{cases} a^2 x + 2 & \text{if } x > 3 \\ 5 & \text{if } x = 3 \\ x^2 - b x + a & \text{if } x < 3 \end{cases}$$ is continuous at $x=3$. 2. **Recall the continuity condition:** For $g(x)$ to be continuous at $x=3$, the left-hand limit, right-hand limit, and the function value at $x=3$ must be equal: $$\lim_{x \to 3^-} g(x) = g(3) = \lim_{x \to 3^+} g(x)$$ 3. **Given limits:** $$\lim_{x \to 3^-} f(x) = 2, \quad \lim_{x \to 3^+} f(x) = -1.5$$ 4. **Apply continuity to $g(x)$:** - Right-hand limit at $x=3$: $$\lim_{x \to 3^+} g(x) = a^2 \cdot 3 + 2 = 3 a^2 + 2$$ - Left-hand limit at $x=3$: $$\lim_{x \to 3^-} g(x) = 3^2 - 3 b + a = 9 - 3 b + a$$ - Function value at $x=3$: $$g(3) = 5$$ 5. **Set limits equal to function value for continuity:** $$3 a^2 + 2 = 5$$ $$9 - 3 b + a = 5$$ 6. **Solve the first equation:** $$3 a^2 + 2 = 5 \implies 3 a^2 = 3 \implies a^2 = 1 \implies a = \pm 1$$ 7. **Solve the second equation for $b$:** $$9 - 3 b + a = 5 \implies -3 b = 5 - 9 - a = -4 - a \implies b = \frac{4 + a}{3}$$ 8. **Substitute $a$ values:** - If $a=1$, then $$b = \frac{4 + 1}{3} = \frac{5}{3}$$ - If $a=-1$, then $$b = \frac{4 - 1}{3} = \frac{3}{3} = 1$$ **Final answer:** $$\boxed{(a,b) = (1, \frac{5}{3}) \text{ or } (-1, 1)}$$