If f(x) = begin{cases} ax + 7, & text{if } x | vedclass

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(D) For $f(x)$ to be continuous at $x = 1$, the condition $\lim_{x 	o 1^-} f(x) = \lim_{x 	o 1^+} f(x) = f(1)$ must hold.Step $1$: Calculate $f(1) =

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$a = -5, b = 2$

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(D) For $f(x)$ to be continuous at $x = 1$, the condition $\lim_{x 	o 1^-} f(x) = \lim_{x 	o 1^+} f(x) = f(1)$ must hold.Step $1$: Calculate $f(1) = 3(1) - 1 = 2$.Step $2$: Calculate the left-hand limit: $\lim_{x 	o 1^-} (ax + 7) = a(1) + 7 = a + 7$. Equating to $f(1)$, we get $a + 7 = 2$, which implies $a = -5$.Step $3$: Calculate the right-hand limit: $\lim_{x 	o 1^+} \frac{x + 3}{b} = \frac{1 + 3}{b} = \frac{4}{b}$. Equating to $f(1)$, we get $\frac{4}{b} = 2$, which implies $b = 2$.Thus, $a = -5$ and $b = 2$.

If $f(x) = \begin{cases} ax + 7, & \text{if } x < 1 \\ 3x - 1, & \text{if } x = 1 \\ \frac{x + 3}{b}, & \text{if } x > 1 \end{cases}$ is continuous at $x = 1$, then

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