To verify whether the given functions [tex]\( f(x) = \frac{x + 5}{2x + 1} \)[/tex] and [tex]\( g(x) = \frac{5 - x}{2x - 1} \)[/tex] are inverses, we need to find [tex]\( g(f(x)) \)[/tex] and confirm if it simplifies to [tex]\( x \)[/tex].
Let's start by finding [tex]\( g(f(x)) \)[/tex]:
1. Determine [tex]\( f(x) \)[/tex]:
[tex]\[
f(x) = \frac{x + 5}{2x + 1}
\][/tex]
2. Substitute [tex]\( f(x) \)[/tex] into [tex]\( g(x) \)[/tex]:
The function [tex]\( g(x) \)[/tex] is defined as:
[tex]\[
g(x) = \frac{5 - x}{2x - 1}
\][/tex]
We need to substitute [tex]\( f(x) \)[/tex] into [tex]\( g(x) \)[/tex]:
[tex]\[
g(f(x)) = g\left(\frac{x + 5}{2x + 1}\right)
\][/tex]
For [tex]\( g(f(x)) \)[/tex], we replace [tex]\( x \)[/tex] in [tex]\( g(x) \)[/tex] with [tex]\( \frac{x + 5}{2x + 1} \)[/tex]:
[tex]\[
g\left(\frac{x + 5}{2x + 1}\right) = \frac{5 - \left(\frac{x + 5}{2x + 1}\right)}{2\left(\frac{x + 5}{2x + 1}\right) - 1}
\][/tex]
3. Simplify the numerator:
[tex]\[
5 - \frac{x + 5}{2x + 1}
\][/tex]
Find a common denominator and subtract:
[tex]\[
5 - \frac{x + 5}{2x + 1} = \frac{5(2x + 1) - (x + 5)}{2x + 1} = \frac{10x + 5 - x - 5}{2x + 1} = \frac{9x}{2x + 1}
\][/tex]
4. Simplify the denominator:
[tex]\[
2\left(\frac{x + 5}{2x + 1}\right) - 1 = \frac{2(x + 5)}{2x + 1} - 1
\][/tex]
Find a common denominator and subtract:
[tex]\[
\frac{2(x + 5) - (2x + 1)}{2x + 1} = \frac{2x + 10 - 2x - 1}{2x + 1} = \frac{9}{2x + 1}
\][/tex]
5. Combine the simplified parts:
Now combining the numerators and the denominators, we get:
[tex]\[
g\left(\frac{x + 5}{2x + 1}\right) = \frac{\frac{9x}{2x + 1}}{\frac{9}{2x + 1}} = x
\][/tex]
Thus, we see that:
[tex]\[
g(f(x)) = x
\][/tex]
This confirms that given functions [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] are indeed inverses of each other.
