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Let's break down the problem step by step:

### Part a:
The surface area \( A(r) \) of a cylinder is given by the equation:

\[
A(r) = 2 \pi r^2 + 8 \pi r
\]

Here, \( r \) is the radius of the cylinder, and we are tasked with finding the domain of \( A(r) \).

#### Step 1: Determine the domain
- The domain of a function is the set of all possible input values (here, \( r \)) for which the function is defined.
- Since the equation contains \( r^2 \) and a linear term \( r \), we need to ensure that \( r \) is non-negative because the radius of a cylinder cannot be negative.
- Therefore, the domain of \( A(r) \) is all \( r \geq 0 \).

**Answer for Part a**: The domain of \( A(r) \) is \( r \geq 0 \).

---

### Part b:
We are asked to solve for \( r \) as a function of \( A \), meaning we need to find the inverse function, \( r(A) \).

The equation we need to solve is:

\[
A = 2 \pi r^2 + 8 \pi r
\]

#### Step 2: Simplifying the equation
Divide both sides by \( 2\pi \) to make the quadratic term easier to manage:

\[
\frac{A}{2\pi} = r^2 + 4r
\]

Now rearrange this as a quadratic equation in standard form:

\[
r^2 + 4r - \frac{A}{2\pi} = 0
\]

#### Step 3: Solve using the quadratic formula
The quadratic formula is:

\[
r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

In our case, \( a = 1 \), \( b = 4 \), and \( c = -\frac{A}{2\pi} \).

Substitute these into the formula:

\[
r = \frac{-4 \pm \sqrt{4^2 - 4(1)\left(-\frac{A}{2\pi}\right)}}{2(1)}
\]

Simplify the discriminant:

\[
r = \frac{-4 \pm \sqrt{16 + 2A/\pi}}{2}
\]

Now divide the terms:

\[
r = \frac{-4 \pm \sqrt{16 + 2A/\pi}}{2}
\]

This gives us two potential solutions. Since the radius must be non-negative, we take the positive root:

\[
r = \frac{-4 + \sqrt{16 + 2A/\pi}}{2}
\]

Thus, the inverse function is:

\[
r(A) = \frac{-4 + \sqrt{16 + 2A/\pi}}{2}
\]

---

### Part c:
Now, we're asked to evaluate \( r \) when \( A = 150 \) square inches.

Substitute \( A = 150 \) into the equation for \( r(A) \):

\[
r(150) = \frac{-4 + \sqrt{16 + 2(150)/\pi}}{2}
\]

I'll calculate the exact value for \( r(150) \).For part c, when the surface area \( A = 150 \) square inches, the radius \( r \) is approximately:

\[
r \approx 3.28 \text{ inches}
\]

Let me know if you'd like further details or have any questions.

Here are 5 related questions to expand your understanding:
1. How would the surface area formula change if the height of the cylinder were different?
2. Can the quadratic formula be used for all values of \( A \), or are there any limitations?
3. How does the domain of the radius function \( r(A) \) change as the surface area increases?
4. What is the interpretation of the negative root in the quadratic formula in this context?
5. How would the result change if the surface area was doubled?

**Tip**: Always check if simplifying the problem (e.g., dividing by common factors like \( 2\pi \)) helps make equations easier to solve!

Assume that the height of your cylinder is 4 inches. Consider A as a function of r, so we can write that as A(r) = 2πr^2 + 8πr. What is the domain of A(r)? In other words, for which values of r is A(r) defined?

This problem explains how to determine the domain of the surface area function A(r) for a cylinder with a height of 4 inches and find the inverse function r(A). It involves the quadratic formula and algebraic manipulation. Ideal for Grades 9-12.

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