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The problem asks to identify the real-valued solutions of \(\frac{f(x)}{g(x)}\), where \(f(x) = x^2 - 16\) and \(g(x) = x^2 - 100\).

Let's begin by solving the problem.

### Step 1: Express \(\frac{f(x)}{g(x)}\)
Given:
\[
f(x) = x^2 - 16
\]
\[
g(x) = x^2 - 100
\]
We want to simplify:
\[
\frac{f(x)}{g(x)} = \frac{x^2 - 16}{x^2 - 100}
\]

### Step 2: Factor both the numerator and denominator
We can factor both the numerator and denominator as differences of squares:
\[
f(x) = x^2 - 16 = (x - 4)(x + 4)
\]
\[
g(x) = x^2 - 100 = (x - 10)(x + 10)
\]
Thus:
\[
\frac{f(x)}{g(x)} = \frac{(x - 4)(x + 4)}{(x - 10)(x + 10)}
\]

### Step 3: Identify the domain
The expression \(\frac{(x - 4)(x + 4)}{(x - 10)(x + 10)}\) is defined for all \(x\) except where the denominator is zero:
\[
x - 10 = 0 \quad \text{or} \quad x + 10 = 0
\]
So, \(x = 10\) and \(x = -10\) are not in the domain of the function.

### Step 4: Simplify and identify the real-valued solutions
The simplified expression \(\frac{(x - 4)(x + 4)}{(x - 10)(x + 10)}\) does not cancel out any terms, so the function remains as is. The real-valued solutions are defined for all \(x\) except \(x = 10\) and \(x = -10\).

Therefore, the expression \(\frac{f(x)}{g(x)}\) can take any real value for all \(x\) except \(x = 10\) and \(x = -10\).

Would you like a detailed explanation of any part of this solution, or do you have any further questions?

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Here are some related questions you might consider:
1. How would you find the roots of the equation \(\frac{f(x)}{g(x)} = 0\)?
2. What happens if the numerator and denominator share a common factor?
3. How would you evaluate \(\frac{f(x)}{g(x)}\) at specific values, such as \(x = 2\)?
4. What is the significance of factoring in simplifying rational expressions?
5. How can you determine if a rational function has vertical asymptotes?

**Tip:** Always check the domain of a rational function by identifying where the denominator equals zero. These points are excluded from the domain and are potential locations for vertical asymptotes.

f(x) = x^2 - 16 and g(x) = x^2 - 100. Identify the real-valued solutions of f(x)/g(x).

Learn how to identify the real-valued solutions of the rational function f(x)/g(x), where f(x) = x^2 - 16 and g(x) = x^2 - 100. This problem covers factoring and determining the domain of rational functions. Suitable for students in Grades 9-11.

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