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Sagot :
To solve the problem of finding the composition [tex]\((g^{-1} \circ f^{-1})(x)\)[/tex] for the given functions [tex]\( f(x) = e^{2x-1} \)[/tex] and [tex]\( g(x) = 2x-1 \)[/tex], let's follow these steps:
Step 1: Find the inverse of [tex]\( f(x) \)[/tex].
Given:
[tex]\[ f(x) = e^{2x-1} \][/tex]
To find the inverse function [tex]\( f^{-1}(x) \)[/tex]:
1. Start by setting [tex]\( y = f(x) \)[/tex]:
[tex]\[ y = e^{2x-1} \][/tex]
2. Solve for [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex]:
[tex]\[ y = e^{2x-1} \][/tex]
Take the natural logarithm on both sides:
[tex]\[ \ln(y) = 2x - 1 \][/tex]
Solve for [tex]\( x \)[/tex]:
[tex]\[ 2x = \ln(y) + 1 \][/tex]
[tex]\[ x = \frac{\ln(y) + 1}{2} \][/tex]
Thus, the inverse function [tex]\( f^{-1}(x) \)[/tex] is:
[tex]\[ f^{-1}(x) = \frac{\ln(x) + 1}{2} \][/tex]
Step 2: Find the inverse of [tex]\( g(x) \)[/tex].
Given:
[tex]\[ g(x) = 2x - 1 \][/tex]
To find the inverse function [tex]\( g^{-1}(x) \)[/tex]:
1. Set [tex]\( y = g(x) \)[/tex]:
[tex]\[ y = 2x - 1 \][/tex]
2. Solve for [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex]:
[tex]\[ y = 2x - 1 \][/tex]
[tex]\[ y + 1 = 2x \][/tex]
[tex]\[ x = \frac{y + 1}{2} \][/tex]
Thus, the inverse function [tex]\( g^{-1}(x) \)[/tex] is:
[tex]\[ g^{-1}(x) = \frac{x + 1}{2} \][/tex]
Step 3: Compute the composition [tex]\((g^{-1} \circ f^{-1})(x)\)[/tex].
To find [tex]\((g^{-1} \circ f^{-1})(x)\)[/tex], we need to substitute [tex]\( f^{-1}(x) \)[/tex] into [tex]\( g^{-1}(x) \)[/tex]:
[tex]\[ (g^{-1} \circ f^{-1})(x) = g^{-1}(f^{-1}(x)) \][/tex]
Substitute [tex]\( f^{-1}(x) \)[/tex] into [tex]\( g^{-1}(x) \)[/tex]:
[tex]\[ g^{-1}\left( \frac{\ln(x) + 1}{2} \right) = \frac{\frac{\ln(x) + 1}{2} + 1}{2} \][/tex]
Simplify the expression:
[tex]\[ \frac{\frac{\ln(x) + 1}{2} + 1}{2} = \frac{\frac{\ln(x) + 1 + 2}{2}}{2} = \frac{\frac{\ln(x) + 3}{2}}{2} = \frac{\ln(x) + 3}{4} \][/tex]
Thus:
[tex]\[ (g^{-1} \circ f^{-1})(x) = \frac{\ln(x) + 3}{4} \][/tex]
Let's compare this with the given options:
(A) [tex]\(\ln x\)[/tex]
(B) [tex]\(e^x\)[/tex]
(C) [tex]\(\frac{1}{4}(1 + \ln(x)) + \frac{1}{2}\)[/tex]:
[tex]\[ = \frac{1}{4}(1 + \ln(x)) + \frac{1}{2} = \frac{1 + \ln(x)}{4} + \frac{2}{4} = \frac{1 + \ln(x) + 2}{4} = \frac{\ln(x) + 3}{4} \][/tex]
(D) [tex]\(\frac{1}{2}\left(1 + \ln\left(\frac{x + 1}{2}\right)\right)\)[/tex]
Hence, the correct answer is (C):
[tex]\[ \boxed{\frac{1}{4}(1+\ln(x))+\frac{1}{2}} \][/tex]
Step 1: Find the inverse of [tex]\( f(x) \)[/tex].
Given:
[tex]\[ f(x) = e^{2x-1} \][/tex]
To find the inverse function [tex]\( f^{-1}(x) \)[/tex]:
1. Start by setting [tex]\( y = f(x) \)[/tex]:
[tex]\[ y = e^{2x-1} \][/tex]
2. Solve for [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex]:
[tex]\[ y = e^{2x-1} \][/tex]
Take the natural logarithm on both sides:
[tex]\[ \ln(y) = 2x - 1 \][/tex]
Solve for [tex]\( x \)[/tex]:
[tex]\[ 2x = \ln(y) + 1 \][/tex]
[tex]\[ x = \frac{\ln(y) + 1}{2} \][/tex]
Thus, the inverse function [tex]\( f^{-1}(x) \)[/tex] is:
[tex]\[ f^{-1}(x) = \frac{\ln(x) + 1}{2} \][/tex]
Step 2: Find the inverse of [tex]\( g(x) \)[/tex].
Given:
[tex]\[ g(x) = 2x - 1 \][/tex]
To find the inverse function [tex]\( g^{-1}(x) \)[/tex]:
1. Set [tex]\( y = g(x) \)[/tex]:
[tex]\[ y = 2x - 1 \][/tex]
2. Solve for [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex]:
[tex]\[ y = 2x - 1 \][/tex]
[tex]\[ y + 1 = 2x \][/tex]
[tex]\[ x = \frac{y + 1}{2} \][/tex]
Thus, the inverse function [tex]\( g^{-1}(x) \)[/tex] is:
[tex]\[ g^{-1}(x) = \frac{x + 1}{2} \][/tex]
Step 3: Compute the composition [tex]\((g^{-1} \circ f^{-1})(x)\)[/tex].
To find [tex]\((g^{-1} \circ f^{-1})(x)\)[/tex], we need to substitute [tex]\( f^{-1}(x) \)[/tex] into [tex]\( g^{-1}(x) \)[/tex]:
[tex]\[ (g^{-1} \circ f^{-1})(x) = g^{-1}(f^{-1}(x)) \][/tex]
Substitute [tex]\( f^{-1}(x) \)[/tex] into [tex]\( g^{-1}(x) \)[/tex]:
[tex]\[ g^{-1}\left( \frac{\ln(x) + 1}{2} \right) = \frac{\frac{\ln(x) + 1}{2} + 1}{2} \][/tex]
Simplify the expression:
[tex]\[ \frac{\frac{\ln(x) + 1}{2} + 1}{2} = \frac{\frac{\ln(x) + 1 + 2}{2}}{2} = \frac{\frac{\ln(x) + 3}{2}}{2} = \frac{\ln(x) + 3}{4} \][/tex]
Thus:
[tex]\[ (g^{-1} \circ f^{-1})(x) = \frac{\ln(x) + 3}{4} \][/tex]
Let's compare this with the given options:
(A) [tex]\(\ln x\)[/tex]
(B) [tex]\(e^x\)[/tex]
(C) [tex]\(\frac{1}{4}(1 + \ln(x)) + \frac{1}{2}\)[/tex]:
[tex]\[ = \frac{1}{4}(1 + \ln(x)) + \frac{1}{2} = \frac{1 + \ln(x)}{4} + \frac{2}{4} = \frac{1 + \ln(x) + 2}{4} = \frac{\ln(x) + 3}{4} \][/tex]
(D) [tex]\(\frac{1}{2}\left(1 + \ln\left(\frac{x + 1}{2}\right)\right)\)[/tex]
Hence, the correct answer is (C):
[tex]\[ \boxed{\frac{1}{4}(1+\ln(x))+\frac{1}{2}} \][/tex]
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