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For $f(x)$ to be continuous at $x=3$ and $x=5$, the left-hand limit (LHL) must equal the right-hand limit (RHL) at these points, and both must equal the function's value at those points.
At $x=3$, we need to ensure that the function is continuous. The left-hand limit (LHL) at $x=3$ is: $$ \lim_{x \to 3^-} f(x) = 1 $$ The right-hand limit (RHL) at $x=3$ is: $$ \lim_{x \to 3^+} f(x) = a(3) + b = 3a + b $$ For continuity at $x=3$, LHL = RHL: $$ 1 = 3a + b $$
At $x=5$, we also need to ensure continuity. The left-hand limit (LHL) at $x=5$ is: $$ \lim_{x \to 5^-} f(x) = a(5) + b = 5a + b $$ The right-hand limit (RHL) at $x=5$ is: $$ \lim_{x \to 5^+} f(x) = 7 $$ For continuity at $x=5$, LHL = RHL: $$ 5a + b = 7 $$
We now have a system of two linear equations: $$ 3a + b = 1 \quad (1) $$ $$ 5a + b = 7 \quad (2) $$ Subtract equation (1) from equation (2): $$ (5a + b) - (3a + b) = 7 - 1 $$ $$ 2a = 6 $$ $$ a = 3 $$ Substitute $a = 3$ into equation (1): $$ 3(3) + b = 1 $$ $$ 9 + b = 1 $$ $$ b = 1 - 9 $$ $$ b = -8 $$
Therefore, $a = 3$ and $b = -8$.
Final Answer: a=3, b=-8