If the function f(x) = begin{cases} 3ax + b, | vedclass

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(D) For a function to be continuous at $x = 1$,the left-hand limit,right-hand limit,and the value of the function at $x = 1$ must be equal.$\lim_{x

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$a = 3, b = 2$

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(D) For a function to be continuous at $x = 1$,the left-hand limit,right-hand limit,and the value of the function at $x = 1$ must be equal.$\lim_{x 	o 1^-} f(x) = \lim_{x 	o 1^-} (3ax + b) = 3a + b$$\lim_{x 	o 1^+} f(x) = \lim_{x 	o 1^+} (5ax - 2b) = 5a - 2b$Given $f(1) = 11$,we have:$3a + b = 11$  (Equation $1$)$5a - 2b = 11$ (Equation $2$)From Equation $1$,$b = 11 - 3a$. Substituting this into Equation $2$:$5a - 2(11 - 3a) = 11$$5a - 22 + 6a = 11$$11a = 33 \implies a = 3$Substituting $a = 3$ into $b = 11 - 3a$:$b = 11 - 3(3) = 11 - 9 = 2$Thus,$a = 3$ and $b = 2$.

If the function $f(x) = \begin{cases} 3ax + b, & \text{for } x < 1 \\ 11, & \text{for } x = 1 \\ 5ax - 2b, & \text{for } x > 1 \end{cases}$ is continuous at $x = 1$,then the values of $a$ and $b$ are:

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