Question

Find the inverse of the function. (Hint: Try rewriting the function by using either inspection or long division.) $$f(x)=\frac{3 x+1}{x-4}$$

Short answer

The inverse of the given function is \(f^{-1}(x) = \frac{4x + 1}{x - 3}\).

Step-by-step solution

Swap x and y

Firstly, rewrite the given function \(f(x)=\frac{3 x+1}{x-4}\) by replacing \(f(x)\) with \(y\), so it becomes \(y=\frac{3 x+1}{x-4}\). Then swap \(x\) and \(y\). This gives us \(x=\frac{3 y+1}{y-4}\). It's done this way because the inverse of a function is found by interchanging the roles of \(x\) and \(y\).

Solve for y

Next step is to isolate \(y\) on one side of the equation. First multiply both sides of the equation by \(y-4\) to get rid of the fraction, yielding \(x(y-4) = 3y + 1\). Distribute \(x\) on the left side to get \(xy-4x = 3y + 1\). Then, try to collect all terms with \(y\) on one side and the remaining terms on the other. This gives us \(xy - 3y = 4x + 1\). At this point, \(y\) can be factorized out on the left side to give \(y(x - 3) = 4x + 1\). Finally, dividing both sides by \(x-3\), we obtain \(y = \frac{4x + 1}{x - 3}\).

Write the inverse function

Now that we've isolated \(y\) on one side, we can write the inverse function. Replace \(y\) with \(f^{-1}(x)\) to get \(f^{-1}(x) = \frac{4x + 1}{x - 3}\). This is the inverse of the original function \(f(x)=\frac{3 x+1}{x-4}\).

Key concepts

Finding Inverse Functions

Understanding how to find the inverse of a function is foundational in Algebra 2, and it is intertwined with the concept of functions themselves. Imagine a function as a machine where you put in an input and get out an output. The inverse function simply reverses this process, accepting the output of the original function as its input and returning the original input as its output.

To find the inverse, we start by replacing the function notation with a simple 'y'. This helps us see the relation in a more familiar 'y equals' format. With the equation set, we then swap 'x' and 'y'. This literally represents putting the output back into the function to get the original input. From the exercise, after swapping, the equation becomes a classic 'solve for y' scenario.

Swapping and Solving for 'y'

Once the 'x' and 'y' are swapped, we face an algebraic puzzle: how to isolate 'y'. This often involves several steps of algebraic manipulation, such as clearing fractions, distributing terms, combining like terms, and factorizing. These steps require careful execution to simplify the equation and solve for 'y'.

Upon solving, the final step in finding the inverse function is to re-substitute 'y' with the inverse function notation, written as 'f^{-1}(x)'. This indicates that the function now takes the place of the original function's outputs and gives back the inputs.

Rational Functions

Rational functions, like the ones in our example, are characterized by the presence of variables in the denominator. They have the form of a fraction where the numerator and the denominator are both polynomials. In the original function, we have a linear polynomial in both the numerator and the denominator, making it a simple rational function.

Rational functions exhibit interesting behaviors such as asymptotes, which are lines that the graph of the function can get arbitrarily close to but never touch. In the function given, we can predict there will be a vertical asymptote at 'x=4' since the function is undefined at that point.

Algebraic Challenges

Manipulating rational functions algebraically can be challenging because it often requires clearing fractions and dealing with polynomial expressions. However, understanding their structure and behaviors can aid in steps such as long division or factoring which are skills crucial in finding their inverses. The bottom line with rational functions is to keep track of conditions, like non-zero denominators, that ensure the expressions are well-defined for real numbers.

Algebraic Manipulation

Algebraic manipulation is the art of reshaping algebraic expressions through a series of mathematically valid moves. When finding inverses, especially of rational functions, these skills are put to the test.

Key Algebraic Processes

For example, to isolate 'y', you may multiply through by the denominator to eliminate fraction form, as seen in the solution to remove the '(y-4)' term. Distributive properties come into play when you need to expand 'x' across '(y-4)', and collecting like terms further simplifies the equation. Finally, factorizing to pull 'y' out of the equation and then dividing sets the stage for solving the problem.

Advanced algebraic manipulation, such as long division or completing the square, may be required in more complex scenarios. Thus, having a strong foundation in these techniques not only helps find inverse functions but is also crucial for tackling a wide variety of algebra problems.