Welcome to Westonci.ca, where curiosity meets expertise. Ask any question and receive fast, accurate answers from our knowledgeable community. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.
Sagot :
We want a function [tex]g[/tex] such that [tex]g(15x-1)=x[/tex]
One easy way to do to that is to choose [tex]g(x)=\dfrac{x+1}{15}[/tex]
That function works : [tex]g(15x-1)=\dfrac{15x-1+1}{15}=x[/tex]
Due to the uniqueness of the inverse, this is the function we were searching for.
One easy way to do to that is to choose [tex]g(x)=\dfrac{x+1}{15}[/tex]
That function works : [tex]g(15x-1)=\dfrac{15x-1+1}{15}=x[/tex]
Due to the uniqueness of the inverse, this is the function we were searching for.
[tex]f(x)=15x-1\to y=15x-1\\\\15x-1=y\ \ \ \ |add\ 1\ to\ both\ sides\\\\15x=y+1\ \ \ \ \ |divide\ both\ sides\ by\ 15\\\\\boxed{x=\frac{y}{15}+\frac{1}{15}}\to\boxed{x=\frac{1}{15}y+\frac{1}{15}}\\\\Answer:\boxed{\boxed{[f(x)]^{-1}=\frac{1}{15}x+\frac{1}{15}}}[/tex]
We hope our answers were helpful. Return anytime for more information and answers to any other questions you may have. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Thank you for trusting Westonci.ca. Don't forget to revisit us for more accurate and insightful answers.