Get the information you need with the help of IDNLearn.com's extensive Q&A platform. Join our community to access reliable and comprehensive responses to your questions from experienced professionals.
Sagot :
To find the inverse of the function [tex]\( f(x) = x + 3 \)[/tex], follow these steps:
1. Rewrite the function as an equation with [tex]\( y \)[/tex]:
[tex]\[ y = x + 3 \][/tex]
2. Interchange the roles of [tex]\( x \)[/tex] and [tex]\( y \)[/tex]: Since we are finding the inverse, we swap [tex]\( x \)[/tex] and [tex]\( y \)[/tex].
[tex]\[ x = y + 3 \][/tex]
3. Solve for [tex]\( y \)[/tex]: Isolate [tex]\( y \)[/tex] on one side of the equation to express it in terms of [tex]\( x \)[/tex].
[tex]\[ y = x - 3 \][/tex]
4. Express the inverse function: Replace [tex]\( y \)[/tex] with [tex]\( h(x) \)[/tex] to denote the inverse function.
[tex]\[ h(x) = x - 3 \][/tex]
So, the inverse of the function [tex]\( f(x) = x + 3 \)[/tex] is:
[tex]\[ h(x) = x - 3 \][/tex]
Therefore, the correct answer is:
[tex]\[ \boxed{h(x) = x - 3} \][/tex]
1. Rewrite the function as an equation with [tex]\( y \)[/tex]:
[tex]\[ y = x + 3 \][/tex]
2. Interchange the roles of [tex]\( x \)[/tex] and [tex]\( y \)[/tex]: Since we are finding the inverse, we swap [tex]\( x \)[/tex] and [tex]\( y \)[/tex].
[tex]\[ x = y + 3 \][/tex]
3. Solve for [tex]\( y \)[/tex]: Isolate [tex]\( y \)[/tex] on one side of the equation to express it in terms of [tex]\( x \)[/tex].
[tex]\[ y = x - 3 \][/tex]
4. Express the inverse function: Replace [tex]\( y \)[/tex] with [tex]\( h(x) \)[/tex] to denote the inverse function.
[tex]\[ h(x) = x - 3 \][/tex]
So, the inverse of the function [tex]\( f(x) = x + 3 \)[/tex] is:
[tex]\[ h(x) = x - 3 \][/tex]
Therefore, the correct answer is:
[tex]\[ \boxed{h(x) = x - 3} \][/tex]
Your participation means a lot to us. Keep sharing information and solutions. This community grows thanks to the amazing contributions from members like you. For dependable answers, trust IDNLearn.com. Thank you for visiting, and we look forward to assisting you again.