Paul Lewis
05/30/2023 · Senior High School
\( \left. \begin{array} { l } { f ( x ) = \frac { 2 } { 2 x + 4 } } \\ { 2 - g ( x ) = \frac { x - 1 } { x ^ { 2 } - x - 2 } } \\ { 3 \cdot h ( x ) = \frac { 3 x + 3 } { 6 x + 9 } } \end{array} \right. \)
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Step-by-step Solution
Let's analyze and simplify each of the functions \( f(x) \), \( g(x) \), and \( h(x) \) given in the problem.
1. **Function \( f(x) \)**:
\[
f(x) = \frac{2}{2x + 4}
\]
We can simplify this by factoring out the common term in the denominator:
\[
f(x) = \frac{2}{2(x + 2)} = \frac{1}{x + 2}
\]
2. **Function \( g(x) \)**:
The equation given is:
\[
2 - g(x) = \frac{x - 1}{x^2 - x - 2}
\]
First, we can factor the denominator:
\[
x^2 - x - 2 = (x - 2)(x + 1)
\]
Thus, we rewrite the equation:
\[
2 - g(x) = \frac{x - 1}{(x - 2)(x + 1)}
\]
To isolate \( g(x) \), we rearrange the equation:
\[
g(x) = 2 - \frac{x - 1}{(x - 2)(x + 1)}
\]
To combine the terms, we express \( 2 \) with a common denominator:
\[
g(x) = \frac{2(x - 2)(x + 1)}{(x - 2)(x + 1)} - \frac{x - 1}{(x - 2)(x + 1)}
\]
This simplifies to:
\[
g(x) = \frac{2(x^2 - x - 2) - (x - 1)}{(x - 2)(x + 1)}
\]
Expanding the numerator:
\[
2(x^2 - x - 2) = 2x^2 - 2x - 4
\]
Thus:
\[
g(x) = \frac{2x^2 - 2x - 4 - x + 1}{(x - 2)(x + 1)} = \frac{2x^2 - 3x - 3}{(x - 2)(x + 1)}
\]
3. **Function \( h(x) \)**:
The equation given is:
\[
3 \cdot h(x) = \frac{3x + 3}{6x + 9}
\]
We can simplify the right side:
\[
3 \cdot h(x) = \frac{3(x + 1)}{3(2x + 3)} = \frac{x + 1}{2x + 3}
\]
Now, to isolate \( h(x) \):
\[
h(x) = \frac{x + 1}{3(2x + 3)} = \frac{x + 1}{6x + 9}
\]
In summary, the simplified functions are:
\[
\begin{align*}
f(x) & = \frac{1}{x + 2}, \\
g(x) & = \frac{2x^2 - 3x - 3}{(x - 2)(x + 1)}, \\
h(x) & = \frac{x + 1}{6x + 9}.
\end{align*}
\]
Quick Answer
\[
\begin{align*}
f(x) & = \frac{1}{x + 2}, \\
g(x) & = \frac{2x^2 - 3x - 3}{(x - 2)(x + 1)}, \\
h(x) & = \frac{x + 1}{6x + 9}.
\end{align*}
\]
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