Find the inverse of the given one-to-one function [tex]f[/tex]. Give the domain and the range of [tex]f[/tex] and of [tex]f^{-1}[/tex], and graph both [tex]f[/tex] and [tex]f^{-1}[/tex] on the same set of axes.
[tex]f(x)=\frac{x+5}{x-3}[/tex]
The inverse function, [tex]f^{-1}(x)=[/tex] [$\square$].
The domain of [tex]f[/tex] is [$\square$].
(Type your answer in interval notation.)
The domain of [tex]f^{-1}[/tex] is [$\square$].
(Type your answer in interval notation.)
The range of [tex]f[/tex] is [$\square$].
(Type your answer in interval notation.)
Answer (2)
Find the inverse function by swapping x and y in the equation y = f r a c x + 5 x − 3 and solving for y , resulting in f − 1 ( x ) = f r a c 3 x + 5 x − 1 .
Determine the domain of f ( x ) by finding the values of x for which the denominator is not zero, giving ( − in f t y , 3 ) c u p ( 3 , in f t y ) .
Find the range of f ( x ) by determining the domain of f − 1 ( x ) , which is ( − in f t y , 1 ) c u p ( 1 , in f t y ) .
State the final answers: f − 1 ( x ) = f r a c 3 x + 5 x − 1 , domain of f is ( − in f t y , 3 ) c u p ( 3 , in f t y ) , domain of f − 1 is ( − in f t y , 1 ) c u p ( 1 , in f t y ) , and range of f is ( − in f t y , 1 ) c u p ( 1 , in f t y ) . f − 1 ( x ) = f r a c 3 x + 5 x − 1
Explanation
Problem Introduction We are given the function f ( x ) = f r a c x + 5 x − 3 and we want to find its inverse, the domain and range of both f and f − 1 .
Finding the Inverse Function To find the inverse function, we first replace f ( x ) with y , so we have y = f r a c x + 5 x − 3 . Next, we swap x and y to get x = f r a c y + 5 y − 3 . Now we solve for y in terms of x .
Solving for y Multiply both sides by ( y − 3 ) to get x ( y − 3 ) = y + 5 . Expanding gives x y − 3 x = y + 5 . Rearranging to isolate y , we have x y − y = 3 x + 5 , so y ( x − 1 ) = 3 x + 5 . Dividing by ( x − 1 ) gives y = f r a c 3 x + 5 x − 1 . Thus, the inverse function is f − 1 ( x ) = f r a c 3 x + 5 x − 1 .
Finding the Domain of f(x) The domain of f ( x ) is all real numbers except where the denominator is zero, so x − 3 n e q 0 , which means x n e q 3 . Thus, the domain of f is ( − in f t y , 3 ) c u p ( 3 , in f t y ) .
Finding the Range of f(x) The range of f ( x ) is the domain of f − 1 ( x ) , which is all real numbers except where the denominator of f − 1 ( x ) is zero, so x − 1 n e q 0 , which means x n e q 1 . Thus, the range of f is ( − in f t y , 1 ) c u p ( 1 , in f t y ) .
Finding the Domain of f^-1(x) The domain of f − 1 ( x ) is all real numbers except where the denominator is zero, so x − 1 n e q 0 , which means x n e q 1 . Thus, the domain of f − 1 is ( − in f t y , 1 ) c u p ( 1 , in f t y ) .
Finding the Range of f^-1(x) The range of f − 1 ( x ) is the domain of f ( x ) , which is all real numbers except where the denominator of f ( x ) is zero, so x − 3 n e q 0 , which means x n e q 3 . Thus, the range of f − 1 is ( − in f t y , 3 ) c u p ( 3 , in f t y ) .
Final Answer Therefore, the inverse function is f − 1 ( x ) = f r a c 3 x + 5 x − 1 , the domain of f is ( − in f t y , 3 ) c u p ( 3 , in f t y ) , the domain of f − 1 is ( − in f t y , 1 ) c u p ( 1 , in f t y ) , and the range of f is ( − in f t y , 1 ) c u p ( 1 , in f t y ) .
Examples
Understanding inverse functions is crucial in many real-world applications. For instance, if you're converting temperatures from Celsius to Fahrenheit using a function, the inverse function allows you to convert back from Fahrenheit to Celsius. Similarly, in cryptography, inverse functions are used to decode messages, ensuring secure communication. These mathematical tools provide a way to reverse processes, which is essential in various scientific and technological fields.
The inverse function of f ( x ) is f − 1 ( x ) = x − 1 3 x + 5 . The domain of f is ( − ∞ , 3 ) ∪ ( 3 , ∞ ) and its range is ( − ∞ , 1 ) ∪ ( 1 , ∞ ) . The domain of f − 1 is ( − ∞ , 1 ) ∪ ( 1 , ∞ ) and its range is ( − ∞ , 3 ) ∪ ( 3 , ∞ ) .
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