To verify whether the functions [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] are inverses of each other, we need to compute [tex]\( f(g(x)) \)[/tex] and see if we get [tex]\( x \)[/tex] as the result. Let's proceed step-by-step:
Given the functions:
[tex]\[ f(x) = \frac{x + 5}{2x + 1} \][/tex]
[tex]\[ g(x) = \frac{5 - x}{2x - 1} \][/tex]
To find [tex]\( f(g(x)) \)[/tex], we need to substitute [tex]\( g(x) \)[/tex] into [tex]\( f(x) \)[/tex]:
[tex]\[ f(g(x)) = \frac{g(x) + 5}{2g(x) + 1} \][/tex]
First, substitute [tex]\( g(x) \)[/tex] into [tex]\( f(x) \)[/tex]:
[tex]\[
f\left(\frac{5 - x}{2x - 1}\right) = \frac{\left(\frac{5 - x}{2x - 1}\right) + 5}{2\left(\frac{5 - x}{2x - 1}\right) + 1}
\][/tex]
Now simplify the numerator and the denominator separately.
Simplifying the Numerator:
[tex]\[
\left(\frac{5 - x}{2x - 1}\right) + 5 = \frac{5 - x + 5(2x - 1)}{2x - 1} = \frac{5 - x + 10x - 5}{2x - 1} = \frac{9x}{2x - 1}
\][/tex]
Simplifying the Denominator:
[tex]\[
2\left(\frac{5 - x}{2x - 1}\right) + 1 = \frac{2(5 - x)}{2x - 1} + 1 = \frac{10 - 2x}{2x - 1} + \frac{2x - 1}{2x - 1} = \frac{10 - 2x + 2x - 1}{2x - 1} = \frac{9}{2x - 1}
\][/tex]
Putting these simplified expressions back into the fraction gives us:
[tex]\[
f(g(x)) = \frac{\frac{9x}{2x - 1}}{\frac{9}{2x - 1}} = \frac{9x}{9} = x
\][/tex]
Hence, we have:
[tex]\[ f(g(x)) = x \][/tex]
This confirms that the functions [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] are indeed inverses of each other.
