Let's break down the provided information step-by-step.
Firstly, we have the given functions:
[tex]\[ f(x) = 15 + x \][/tex]
[tex]\[ g(x) = x - 3 \][/tex]
We need to define the function [tex]\( h(x) \)[/tex] as the division of [tex]\( f(x) \)[/tex] by [tex]\( g(x) \)[/tex]:
[tex]\[ h(x) = \frac{f(x)}{g(x)} = \frac{15 + x}{x - 3} \][/tex]
Given that we want [tex]\( h(x) = \frac{15 + x}{x - 3} \)[/tex], we observe that this fraction does not directly align with any predefined simple form like [tex]\( x - 5 \)[/tex]. To fit the context of the question effectively, we skip this specific pattern matching and instead focus on calculating the value and domain.
The earlier results indicate that the value of [tex]\( h(x) \)[/tex] at [tex]\( x = 8 \)[/tex] is calculated as [tex]\( 4.6 \)[/tex]. Implies that the interpretation is correct.
Next, we consider the domain. We know that a rational function (a ratio of two polynomials) like [tex]\( h(x) \)[/tex] is defined wherever the denominator isn't zero. So, we set [tex]\( g(x) = 0 \)[/tex] to find where it is undefined:
[tex]\[ x - 3 = 0 \implies x = 3 \][/tex]
Thus, the function [tex]\( h(x) = \frac{f(x)}{g(x)} \)[/tex] is undefined at [tex]\( x = 3 \)[/tex]. This finding divides the set of all real numbers into two intervals:
[tex]\[ (-\infty, 3) \quad \text{and} \quad (3, \infty) \][/tex]
These two intervals represent the domain of [tex]\( h(x) \)[/tex].
By combining all this information:
[tex]\[ h(x) = \frac{15 + x}{x - 3} \][/tex]
The domain of [tex]\( h(x) \)[/tex] is [tex]\((- \infty, 3) \cup (3, \infty)\)[/tex].
Thus, the answers are:
[tex]\[ h(x) = \frac{15 + x}{x - 3} \][/tex]
The domain of [tex]\( h(x) \)[/tex] is [tex]\((- \infty, 3) \cup (3, \infty)\)[/tex].
