Let f(t) = $\frac{t + 3}{t^2 - 5t - 66}$

Now, to find the domain, we first factor the denominator of the function f (t). The denominator of function f(t) is t

^{2}- 5t - 66.

To factor this, we divide -5t, such that addition becomes -5t and multiplication becomes -66. So,

t

^{2}- 11t + 5t - 66 = 0

t(t - 11) + 5(t - 11) = 0

(t - 11) (t + 5) = 0

So, the term t

^{2 }- 5t - 66 contains two factors (t - 11) and (t + 5). Here, the domain of function f (t) is all values of ‘t’ which can be substituted and answer is always defined.

So, denominator = (t - 12) (t + 2). So, the domain of function f (t) = $(- \infty, -2) U (-2, 11) U (11, \infty)$.

Now, we have to find out the range.

So, f (t) = $\frac{t + 3}{t^2 - 5t - 66}$

y = $\frac{t + 3}{t^2 - 5t - 66}$

After that, we multiply both the sides by the Least Common Denominator,

y(t

^{2}- 5t - 66) = (t + 3)

yt

^{2 }- 5yt - t - 66y - 3 = 0

yt

^{2}- (5y + 1) t - 66y - 3 = 0

yt

^{2}- (5y + 1) t - 3(22y + 1) = 0

Resulting equation is looking like Quadratic Equation and the range of all real numbers is $(-\infty, \infty)$. So, the range of the function f (t) is also $(-\infty, \infty)$.