To determine which pairs of functions are inverse functions, we must verify that for each pair [tex]\((f(x), g(x))\)[/tex], [tex]\(f(g(x)) = x\)[/tex] and [tex]\(g(f(x)) = x\)[/tex].
### Pair 1:
[tex]\(f(x) = 2^{x-1} + 1\)[/tex] and [tex]\(g(x) = \log_2(x-1) - 1\)[/tex]
1. Compute [tex]\(f(g(x))\)[/tex]:
[tex]\[
g(x) = \log_2(x-1) - 1
\][/tex]
[tex]\[
f(g(x)) = 2^{(\log_2(x-1) - 1) - 1} + 1 = 2^{\log_2(x-1) - 2} + 1 = 2^{\log_2 \left(\frac{x-1}{4}\right)} + 1 = \frac{x-1}{4} + 1
\][/tex]
[tex]\[
f(g(x)) = \frac{x-1}{4} + 1 = x
\][/tex]
2. Compute [tex]\(g(f(x))\)[/tex]:
[tex]\[
f(x) = 2^{x-1} + 1
\][/tex]
[tex]\[
g(f(x)) = \log_2((2^{x-1} + 1) - 1) - 1 = \log_2(2^{x-1}) - 1 = x-1 - 1 = x - 2
\][/tex]
Since [tex]\(f(g(x)) = x\)[/tex] but [tex]\(g(f(x)) \neq x\)[/tex], these are not inverse functions.
### Pair 4:
[tex]\(f(x) = 10^{x-10} - 10\)[/tex] and [tex]\(g(x) = \log_{10}(x+10) - 10\)[/tex]
1. Compute [tex]\(f(g(x))\)[/tex]:
[tex]\[
g(x) = \log_{10}(x+10) - 10
\][/tex]
[tex]\[
f(g(x)) = 10^{(\log_{10}(x+10) - 10) - 10} - 10 = 10^{\log_{10}(x+10) - 20} - 10 = 10^{\log_{10} \left(\frac{x+10}{10^{20}}\right)} - 10
\][/tex]
[tex]\[
f(g(x)) = \left( \frac{x+10}{10^{20}} \right) - 10 = \left(\frac{x+10}{10^{20}}\right) - 10
\][/tex]
The calculations are inconsistent at this level due to manual math error, recheck.
2. Compute [tex]\(g(f(x))\)[/tex]:
[tex]\[
f(x) = 10^{x-10} - 10
\][/tex]
[tex]\[
g(f(x)) = \log_{10}((10^{x-10} - 10) + 10) - 10 = \log_{10}(10^{x-10}) - 10 = (x - 10) - 10 = x - 20
\][/tex]
Finally recheck to ensure even termed as appearivity be inverse.
