To determine the domain of the function [tex]\(\left(\frac{f}{g}\right)(x)\)[/tex] where [tex]\(f(x) = x^2 - 25\)[/tex] and [tex]\(g(x) = x - 5\)[/tex], we need to find the values of [tex]\(x\)[/tex] for which the function is defined.
1. Consider the denominator [tex]\(g(x)\)[/tex] of the function first:
[tex]\[
g(x) = x - 5
\][/tex]
For the function [tex]\(\left(\frac{f}{g}\right)(x)\)[/tex] to be defined, the denominator [tex]\(g(x)\)[/tex] should not be zero. Setting the denominator equal to zero to find the critical points:
[tex]\[
x - 5 = 0
\][/tex]
Solving for [tex]\(x\)[/tex]:
[tex]\[
x = 5
\][/tex]
So, the function [tex]\(\left(\frac{f}{g}\right)(x)\)[/tex] is not defined at [tex]\(x = 5\)[/tex].
2. Consider the numerator [tex]\(f(x)\)[/tex]:
[tex]\[
f(x) = x^2 - 25
\][/tex]
Factor it to better identify any restrictions:
[tex]\[
f(x) = (x + 5)(x - 5)
\][/tex]
The factored form shows that [tex]\(f(x)\)[/tex] has zeros at:
[tex]\[
x = 5 \quad \text{and} \quad x = -5
\][/tex]
However, the zeros of [tex]\(f(x)\)[/tex] do not affect the domain of [tex]\(\left(\frac{f}{g}\right)(x)\)[/tex] except through [tex]\(g(x)\)[/tex].
3. Determine the actual restrictions on the domain:
- The only value that makes the denominator zero is [tex]\(x = 5\)[/tex].
Hence, the domain of the function [tex]\(\left(\frac{f}{g}\right)(x)\)[/tex] is all real values of [tex]\(x\)[/tex] except where [tex]\(x = 5\)[/tex].
So, the correct answer is:
[tex]\[ \text{all real values of } x \text{ except } x=5 \][/tex]
