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For the function to be continuous at $x = 2$, the left-hand limit must equal the right-hand limit: $$ \lim_{x \to 2^-} (\frac{1}{2}x^2 + 2x) = \lim_{x \to 2^+} (ax + b) $$ $$ \frac{1}{2}(2)^2 + 2(2) = a(2) + b $$ $$ 2 + 4 = 2a + b \implies 2a + b = 6 $$
The derivative of the first segment is $h'(x) = \frac{d}{dx}(\frac{1}{2}x^2 + 2x) = x + 2$. Evaluating at $x = 2$: $$ h'(2^-) = 2 + 2 = 4 $$
For smoothness, the derivative of the second segment must match the left-hand derivative at $x = 2$. The derivative of the second segment is $h'(x) = a$. Setting $a = 4$: Substitute $a = 4$ into the continuity equation $2a + b = 6$: $$ 2(4) + b = 6 \implies 8 + b = 6 \implies b = -2 $$ Thus, $a = 4$ and $b = -2$.
Final Answer: (i) 2a + b = 6; (ii) 4; (iii) a = 4, b = -2
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