Certainly! Let's define the functions [tex]\( f \)[/tex] and [tex]\( g \)[/tex]:
[tex]\[ f(x) = x - 5 \][/tex]
[tex]\[ g(x) = x^2 - 1 \][/tex]
We will find the following expressions step-by-step using an arbitrary value [tex]\( x = 2 \)[/tex]:
1. [tex]\( f + g \)[/tex]:
[tex]\[
(f + g)(x) = f(x) + g(x)
\][/tex]
Given [tex]\( x = 2 \)[/tex]:
[tex]\[
f(2) = 2 - 5 = -3
\][/tex]
[tex]\[
g(2) = 2^2 - 1 = 4 - 1 = 3
\][/tex]
[tex]\[
(f + g)(2) = f(2) + g(2) = -3 + 3 = 0
\][/tex]
Therefore, [tex]\( (f + g)(2) = 0 \)[/tex].
2. [tex]\( f - g \)[/tex]:
[tex]\[
(f - g)(x) = f(x) - g(x)
\][/tex]
Given [tex]\( x = 2 \)[/tex]:
[tex]\[
f(2) = -3
\][/tex]
[tex]\[
g(2) = 3
\][/tex]
[tex]\[
(f - g)(2) = f(2) - g(2) = -3 - 3 = -6
\][/tex]
Therefore, [tex]\( (f - g)(2) = -6 \)[/tex].
3. [tex]\( f \cdot g \)[/tex]:
[tex]\[
(f \cdot g)(x) = f(x) \cdot g(x)
\][/tex]
Given [tex]\( x = 2 \)[/tex]:
[tex]\[
f(2) = -3
\][/tex]
[tex]\[
g(2) = 3
\][/tex]
[tex]\[
(f \cdot g)(2) = f(2) \cdot g(2) = -3 \cdot 3 = -9
\][/tex]
Therefore, [tex]\( (f \cdot g)(2) = -9 \)[/tex].
4. [tex]\( \frac{f}{g} \)[/tex]:
[tex]\[
\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)}
\][/tex]
Given [tex]\( x = 2 \)[/tex]:
[tex]\[
f(2) = -3
\][/tex]
[tex]\[
g(2) = 3
\][/tex]
[tex]\[
\left(\frac{f}{g}\right)(2) = \frac{f(2)}{g(2)} = \frac{-3}{3} = -1.0
\][/tex]
Therefore, [tex]\( \frac{f}{g}(2) = -1.0 \)[/tex].
5. [tex]\( \frac{g}{g} \)[/tex]:
[tex]\[
\left(\frac{g}{g}\right)(x) = \frac{g(x)}{g(x)}
\][/tex]
Given [tex]\( x = 2 \)[/tex]:
[tex]\[
g(2) = 3
\][/tex]
[tex]\[
\left(\frac{g}{g}\right)(2) = \frac{g(2)}{g(2)} = \frac{3}{3} = 1.0
\][/tex]
Therefore, [tex]\( \frac{g}{g}(2) = 1.0 \)[/tex].
To summarize, when [tex]\( x = 2 \)[/tex]:
[tex]\[
\begin{align*}
(f + g)(2) &= 0 \\
(f - g)(2) &= -6 \\
(f \cdot g)(2) &= -9 \\
\left(\frac{f}{g}\right)(2) &= -1.0 \\
\left(\frac{g}{g}\right)(2) &= 1.0 \\
\end{align*}
\][/tex]
