A) To derive the function that relates oral dosage to the number of surviving bacteria, we compose the given functions.

a) To derive the function that relates oral dosage to the number of surviving bacteria, we compose the given functions.
Let $D$ be the oral dosage in mg.
The amount of antibiotic entering the bloodstream is $h(D) = \frac{1}{2}D$.
The amount absorbed into the site of infection from the bloodstream is $g(h(D))$.
The number of surviving bacteria is $f(g(h(D)))$.

Step 1: Compose $g(h(D))$.

$$g(h(D)) = g\left(\frac{1}{2}D\right) = \frac{\left(\frac{1}{2}D\right)+4}{4\left(\frac{1}{2}D\right)}$$ $$g(h(D)) = \frac{\frac{D}{2}+4}{2D} = \frac{\frac{D+8}{2}}{2D} = \frac{D+8}{4D}$$

This is the amount of antibiotic absorbed into the site of infection for an oral dosage $D$.

Step 2: Compose $f(g(h(D)))$.
Let $A_{oral}(D) = \frac{D+8}{4D}$ be the amount absorbed at the infection site.

$$f(A_{oral}(D)) = f\left(\frac{D+8}{4D}\right) = \frac{3200}{8+\left(\frac{D+8}{4D}\right)^2}$$

The function relating oral dosage $D$ to the number of surviving bacteria is:

$$\boxed{S_{oral}(D) = \frac{3200}{8+\left(\frac{D+8}{4D}\right)^2}}$$

b) If the antibiotics are administered by injection, we assume the injected dosage $D_{inj}$ directly enters the bloodstream (100% bioavailability).
The amount absorbed into the site of infection is $g(D_{inj})$.
The number of surviving bacteria is $f(g(D_{inj}))$.

Step 1: Compose $f(g(D_{inj}))$.

$$f(g(D_{inj})) = f\left(\frac{D_{inj}+4}{4D_{inj}}\right) = \frac{3200}{8+\left(\frac{D_{inj}+4}{4D_{inj}}\right)^2}$$

The function relating injected dosage $D_{inj}$ to the number of surviving bacteria is:

$$\boxed{S_{inj}(D_{inj}) = \frac{3200}{8+\left(\frac{D_{inj}+4}{4D_{inj}}\right)^2}}$$

c) To determine which treatment is best for a 3 mg dosage, we calculate the number of surviving bacteria for both oral and injected treatments. The best treatment will result in the lowest number of surviving bacteria.

Step 1: Calculate surviving bacteria for oral dosage of 3 mg.
Substitute $D=3$ into $S_{oral}(D)$:

$$S_{oral}(3) = \frac{3200}{8+\left(\frac{3+8}{4 \times 3}\right)^2} = \frac{3200}{8+\left(\frac{11}{12}\right)^2}$$ $$S_{oral}(3) = \frac{3200}{8+\frac{121}{144}} = \frac{3200}{\frac{8 \times 144 + 121}{144}} = \frac{3200}{\frac{1152+121}{144}}$$ $$$$