At Westonci.ca, we make it easy to get the answers you need from a community of informed and experienced contributors. Get immediate and reliable solutions to your questions from a community of experienced experts on our Q&A platform. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.
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]
We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Westonci.ca is your go-to source for reliable answers. Return soon for more expert insights.