1. **State the problem:** Find values of $a$ and $b$ such that the piecewise function $$ f(x) = \begin{cases} \frac{x^2 - 4}{x - 2} & x < 2 \\ ax^2 - bx + 3 & 2 \leq x < 3 \\ 2x - a + b & x \geq 3 \end{cases} $$ is continuous everywhere. 2. **Continuity conditions:** For $f$ to be continuous everywhere, it must be continuous at the points where the definition changes, i.e., at $x=2$ and $x=3$. 3. **At $x=2$ continuity:** - Left limit as $x \to 2^-$: $$ \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} \frac{x^2 - 4}{x - 2} = \lim_{x \to 2^-} \frac{(x-2)(x+2)}{x-2}. $$ - Cancel common factor: $$ \frac{\cancel{(x-2)}(x+2)}{\cancel{(x-2)}} = x + 2. $$ - Evaluate limit: $$ \lim_{x \to 2^-} (x+2) = 2 + 2 = 4. $$ - Right limit as $x \to 2^+$: $$ \lim_{x \to 2^+} f(x) = a(2)^2 - b(2) + 3 = 4a - 2b + 3. $$ - For continuity at $x=2$: $$ 4 = 4a - 2b + 3 \implies 4a - 2b = 1. $$ 4. **At $x=3$ continuity:** - Left limit as $x \to 3^-$: $$ \lim_{x \to 3^-} f(x) = a(3)^2 - b(3) + 3 = 9a - 3b + 3. $$ - Right limit as $x \to 3^+$: $$ \lim_{x \to 3^+} f(x) = 2(3) - a + b = 6 - a + b. $$ - For continuity at $x=3$: $$ 9a - 3b + 3 = 6 - a + b \implies 10a - 4b = 3. $$ 5. **System of equations from continuity:** $$ \begin{cases} 4a - 2b = 1 \\ 10a - 4b = 3 \end{cases} $$ 6. **Given relations:** $$ a = 3 + \frac{3b}{5}, \quad b = -1 + \frac{5a}{3}. $$ 7. **Solve system for $a$ and $b$:** - From first equation: $$ 4a - 2b = 1 \implies 2a - b = \frac{1}{2}. $$ - Express $b$: $$ b = 2a - \frac{1}{2}. $$ - Substitute into second equation: $$ 10a - 4b = 3 \implies 10a - 4(2a - \frac{1}{2}) = 3. $$ - Simplify: $$ 10a - 8a + 2 = 3 \implies 2a + 2 = 3 \implies 2a = 1 \implies a = \frac{1}{2}. $$ - Find $b$: $$ b = 2 \times \frac{1}{2} - \frac{1}{2} = 1 - \frac{1}{2} = \frac{1}{2}. $$ 8. **Check given relations:** - Check $a = 3 + \frac{3b}{5}$: $$ 3 + \frac{3 \times \frac{1}{2}}{5} = 3 + \frac{3/2}{5} = 3 + \frac{3}{10} = 3.3 \neq \frac{1}{2}. $$ - Check $b = -1 + \frac{5a}{3}$: $$ -1 + \frac{5 \times \frac{1}{2}}{3} = -1 + \frac{5/2}{3} = -1 + \frac{5}{6} = -\frac{1}{6} \neq \frac{1}{2}. $$ The given relations contradict the continuity conditions. The values $a=\frac{1}{2}$ and $b=\frac{1}{2}$ satisfy continuity but not the given relations. **Final answer:** $$ a = \frac{1}{2}, \quad b = \frac{1}{2}. $$