To solve for the inverse function [tex]\( f^{-1}(x) \)[/tex] of the given function [tex]\( f(x) \)[/tex], let's examine the function [tex]\( f(x) \)[/tex] closely.
First, let's clarify the given function. The function stated is:
[tex]\[ f(x) = 10^2 + 1 \][/tex]
In this form, [tex]\( f(x) \)[/tex] is a constant function because it does not contain the variable [tex]\( x \)[/tex] at all. Calculating the expression:
[tex]\[ f(x) = 10^2 + 1 = 100 + 1 = 101 \][/tex]
So, [tex]\( f(x) = 101 \)[/tex]. This means that for any value of [tex]\( x \)[/tex], the output is always 101.
However, typically when asking for the inverse of a function, we usually want to find a function [tex]\( g(y) \)[/tex] such that if [tex]\( y = f(x) \)[/tex], then [tex]\( x = g(y) \)[/tex]. Since [tex]\( f(x) \)[/tex] is a constant function and doesn't change with [tex]\( x \)[/tex], it doesn't have an inverse function in the usual sense. An inverse function essentially "reverses" the effect of the original function.
But let's assume there was a mistake or ambiguity in the problem statement and that the intended function was actually:
[tex]\[ f(x) = 10^x + 1 \][/tex]
In this case, we can proceed to find the inverse function:
1. Start with [tex]\( y = f(x) \)[/tex]:
[tex]\[ y = 10^x + 1 \][/tex]
2. Subtract 1 from both sides to isolate the exponential term:
[tex]\[ y - 1 = 10^x \][/tex]
3. Take the logarithm base 10 of both sides to solve for [tex]\( x \)[/tex]:
[tex]\[ \log_{10}(y - 1) = x \][/tex]
So the inverse function, [tex]\( f^{-1}(x) \)[/tex], in terms of [tex]\( x \)[/tex], is:
[tex]\[ f^{-1}(x) = \log_{10}(x - 1) \][/tex]
Therefore, the correct answer is:
[tex]\[ f^{-1}(x) = \log (x - 1) \][/tex]
