### Part a: Determining the Domain of [tex]\( A(r) \)[/tex]
Consider the function [tex]\( A(r) \)[/tex] representing the surface area of a cylinder with height 4 inches:
[tex]\[ A(r) = 2 \pi r^2 + 8 \pi r \][/tex]
To determine the domain, we need to find the values of [tex]\( r \)[/tex] for which this function is defined. Since [tex]\( r \)[/tex] represents the radius of a cylinder, it cannot be negative. Therefore, the radius [tex]\( r \)[/tex] must be greater than or equal to zero:
[tex]\[ r \geq 0 \][/tex]
Thus, the domain of [tex]\( A(r) \)[/tex] is:
[tex]\[ \boxed{(0, \infty)} \][/tex]
### Part b: Finding the Inverse Function [tex]\( r(A) \)[/tex]
To find the inverse function [tex]\( r(A) \)[/tex], we start with the equation for the surface area and solve for [tex]\( r \)[/tex]:
[tex]\[ A = 2 \pi r^2 + 8 \pi r \][/tex]
Rewriting it in standard form:
[tex]\[ 2 \pi r^2 + 8 \pi r - A = 0 \][/tex]
This is a quadratic equation in the variable [tex]\( r \)[/tex]. Using the quadratic formula [tex]\( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex] with:
[tex]\[ a = 2 \pi, \quad b = 8 \pi, \quad c = -A \][/tex]
We have:
[tex]\[ r = \frac{-8 \pi \pm \sqrt{(8 \pi)^2 - 4 \cdot 2 \pi \cdot (-A)}}{2 \cdot 2 \pi} \][/tex]
Simplifying the expression inside the square root:
[tex]\[ r = \frac{-8 \pi \pm \sqrt{64 \pi^2 + 8 \pi A}}{4 \pi} \][/tex]
Taking the [tex]\( \pi \)[/tex] common inside the square root:
[tex]\[ r = \frac{-8 \pm \sqrt{64 + 8A/\pi}}{4} \][/tex]
Factoring out common terms:
[tex]\[ r = \frac{-8 \pm 2 \sqrt{16 + 2A/\pi}}{4} \][/tex]
[tex]\[ r = \frac{-8}{4} \pm \frac{2 \sqrt{16 + 2A/\pi}}{4} \][/tex]
[tex]\[ r = -2 \pm \frac{\sqrt{16 + 2A/\pi}}{2} \][/tex]
Simplifying further:
[tex]\[ r = -2 \pm \sqrt{0.5 \frac{A}{\pi} + 4} \][/tex]
Therefore, the two solutions for [tex]\( r \)[/tex] as functions of [tex]\( A \)[/tex] are:
[tex]\[ r(A) = -2 - \sqrt{0.5 \frac{A}{\pi} + 4} \][/tex]
[tex]\[ r(A) = -2 + \sqrt{0.5 \frac{A}{\pi} + 4} \][/tex]
Since [tex]\( r \)[/tex] must be non-negative, we take the positive root:
[tex]\[ r(A) = -2 + \sqrt{0.5 \frac{A}{\pi} + 4} \][/tex]
### Part c: Evaluating [tex]\( r(200) \)[/tex]
To find the radius when the surface area is 200 square inches, we substitute [tex]\( A = 200 \)[/tex] into the inverse function:
[tex]\[ r(200) = -2 + \sqrt{0.5 \frac{200}{\pi} + 4} \][/tex]
Plugging in the values:
[tex]\[ r(200) = -2 + \sqrt{0.5 \cdot \frac{200}{\pi} + 4} \][/tex]
[tex]\[ r(200) = -2 + \sqrt{0.5 \cdot \left(\frac{200}{3.141592653589793}\right) + 4} \][/tex]
[tex]\[ r(200) \approx -2 + \sqrt{31.831 + 4} \][/tex]
[tex]\[ r(200) = -2 + \sqrt{35.831} \][/tex]
[tex]\[ r(200) \approx -2 + 5.9859 \][/tex]
[tex]\[ r(200) \approx 3.99 \][/tex]
Thus, the radius [tex]\( r \)[/tex] when the surface area is 200 square inches is approximately:
[tex]\[ \boxed{3.99} \][/tex]
