To determine the domain and range of the function [tex]\(f(x) = \frac{x^2 + x + 8}{x^2 - 10x + 16}\)[/tex], we can follow these steps carefully.
### Step 1: Determine the Domain
The domain of [tex]\(f(x)\)[/tex] includes all real numbers [tex]\(x\)[/tex] for which the denominator is not zero. Therefore, we need to solve:
[tex]\[
x^2 - 10x + 16 = 0
\][/tex]
Solve this quadratic equation using the quadratic formula, [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 1\)[/tex], [tex]\(b = -10\)[/tex], and [tex]\(c = 16\)[/tex]:
[tex]\[
x = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 1 \cdot 16}}{2 \cdot 1} = \frac{10 \pm \sqrt{100 - 64}}{2} = \frac{10 \pm \sqrt{36}}{2} = \frac{10 \pm 6}{2}
\][/tex]
This gives us two solutions:
[tex]\[
x = \frac{10 + 6}{2} = 8 \quad \text{and} \quad x = \frac{10 - 6}{2} = 2
\][/tex]
So, the domain of [tex]\(f(x)\)[/tex] is all real numbers except [tex]\(x = 2\)[/tex] and [tex]\(x = 8\)[/tex]. We can write the domain as:
[tex]\[
\text{Domain} = \mathbb{R} \setminus \{2, 8\}
\][/tex]
### Step 2: Determine the Range
The range requires us to analyze the values that [tex]\(f(x)\)[/tex] can take. To find the possible values of [tex]\(y\)[/tex] such that [tex]\(y = f(x)\)[/tex], we need to set up the equation [tex]\(f(x) = y\)[/tex] and solve for [tex]\(x\)[/tex]:
[tex]\[
y = \frac{x^2 + x + 8}{x^2 - 10x + 16}
\][/tex]
Rewriting this in a more manageable form:
[tex]\[
y(x^2 - 10x + 16) = x^2 + x + 8 \quad \Rightarrow \quad yx^2 - 10yx + 16y = x^2 + x + 8
\][/tex]
By rearranging terms, we get a quadratic equation in [tex]\(x\)[/tex]:
[tex]\[
(y - 1)x^2 + (10y + 1)x + (16y - 8) = 0
\][/tex]
For specific values of [tex]\(y\)[/tex], the quadratic must have real solutions for [tex]\(x\)[/tex]. However, this doesn't guarantee that all values of [tex]\(y\)[/tex] yield valid solutions that lie within our domain restrictions. Upon closer inspection and solving the equation, we find that the values of [tex]\(y\)[/tex] that make the denominator zero are not valid for the function.
Given this, we can summarize our solution:
[tex]\[
\text{Range} = \emptyset
\][/tex]
### Conclusion:
Therefore, the domain of the function [tex]\(f(x) = \frac{x^2 + x + 8}{x^2 - 10x + 16}\)[/tex] is:
[tex]\[
\mathbb{R} \setminus \{2, 8\}
\][/tex]
and the range of the function is:
[tex]\[
\emptyset
\][/tex]
Since the function exhibits complexity in solving the quadratic for real solutions within the defined domain, we conclude that no valid range exists within the operational constraints.
