Passage: Let S and T be the sets where...

Given $f(x) = \dfrac{x^3}{3} - \dfrac{5x^2}{2} + 6x + 7$

Differentiating with respect to $x$, we get:

$f'(x) = x^2 - 5x + 6$

$f'(x) = (x - 2)(x - 3)$

For the function to be increasing, $f'(x) \geq 0$

$(x - 2)(x - 3) \geq 0$

This inequality holds true for $x \leq 2$ or $x \geq 3$.

Therefore, the set $T$ where the function increases is $\{x \leq 2\} \cup \{x \geq 3\}$.

Answer: $\{x\leq2\}\cup\{x\geq3\}$