Passage: Let f(x)= cases ax(x-1), & x 3 cases Given that f(x) is...

For $f(x)$ to be continuous at $x=3$, we must have $\lim_{x \to 3^-} f(x) = \lim_{x \to 3^+} f(x)$.

$3 - 1 = p(3)^2 + q(3) + 2$

$2 = 9p + 3q + 2$

$9p + 3q = 0 \Rightarrow 3p + q = 0$

For $f'(x)$ to be continuous at $x=3$, the left-hand derivative must equal the right-hand derivative at $x=3$.

$f'(x) = 1$ for $1 < x < 3$

$f'(x) = 2px + q$ for $x > 3$

Equating the derivatives at $x=3$:

$1 = 2p(3) + q$

$6p + q = 1$

Solving the two equations:

$3p + q = 0$

$6p + q = 1$

Subtracting the first equation from the second gives $3p = 1 \Rightarrow p = \dfrac{1}{3}$.

Substituting $p$ into the first equation gives $q = -3\left(\dfrac{1}{3}\right) = -1$.

Answer: $-1$