A function is defined as f(x) = (72)/(px+r). Given f(6)=12 and f(7)=9, find: a) the values of p and r. b) f⁻¹((5)/(3))

Given $f(6) = 12$:

$$12 = \frac{72}{p(6)+r}$$ $$12(6p+r) = 72$$ $$6p+r = \frac{72}{12}$$ $$6p+r = 6 \quad (\text{Equation 1})$$

Given $f(7) = 9$:

$$9 = \frac{72}{p(7)+r}$$ $$9(7p+r) = 72$$ $$7p+r = \frac{72}{9}$$ $$7p+r = 8 \quad (\text{Equation 2})$$

Subtract Equation 1 from Equation 2:

$$(7p+r) - (6p+r) = 8 - 6$$ $$p = 2$$

Substitute $p=2$ into Equation 1:

$$6(2)+r = 6$$ $$12+r = 6$$ $$r = 6-12$$ $$r = -6$$

The values are $p = \boxed{2}$ and $r = \boxed{-6}$.

Using $p=2$ and $r=-6$, the function is $f(x) = \frac{72}{2x-6}$.
Let $y = f(x)$. So, $y = \frac{72}{2x-6}$.
To find the inverse, swap $x$ and $y$:

$$x = \frac{72}{2y-6}$$

Solve for $y$:

$$x(2y-6) = 72$$ $$2y-6 = \frac{72}{x}$$ $$2y = \frac{72}{x} + 6$$ $$y = \frac{36}{x} + 3$$

So, $f^{-1}(x) = \frac{36}{x} + 3$.

Substitute $x = \frac{5}{3}$ into the inverse function:

$$f^{-1}\left(\frac{5}{3}\right) = \frac{36}{\frac{5}{3}} + 3$$ $$f^{-1}\left(\frac{5}{3}\right) = 36 \times \frac{3}{5} + 3$$ $$f^{-1}\left(\frac{5}{3}\right) = \frac{108}{5} + 3$$ $$f^{-1}\left(\frac{5}{3}\right) = \frac{108}{5} + \frac{15}{5}$$ $$f^{-1}\left(\frac{5}{3}\right) = \frac{123}{5}$$

The value is $f^{-1}\left(\frac{5}{3}\right) = \boxed{\frac{123}{5}}$.