Problem 12 - Precalculus with Limits

Question

In Exercises 7-14, find the inverse function of \(f\) informally. Verify that \(f(f^{-1}(x)) = x\) and \(f^{-1}(f(x)) = x\). \(f(x) = \frac{x-1}{5}\)

Step-by-Step Solution

Answer

The inverse function of \(f(x) = \frac{x-1}{5}\) is \(f^{-1}(x) = 5x+1\). The verifications \(f(f^{-1}(x)) = x\) and \(f^{-1}(f(x)) = x\) have been proven to be true.

Step 1: Finding the Inverse Function \(f^{-1}(x)\)

To find the inverse function \(f^{-1}(x)\), start by re-writing the original function \(f(x) = \frac{x-1}{5}\) with 'y' instead of 'f(x)' to get \(y = \frac{x-1}{5}\). Then, swap 'x' and 'y' to get \(x = \frac{y-1}{5}\). Now solve for 'y' to derive the inverse function: multiplied both sides by 5, then add 1 to both sides, the result is \(f^{-1}(x) = 5x+1\).

Step 2: Verifying the first equation \(f(f^{-1}(x)) = x\)

To verify that \(f(f^{-1}(x)) = x\), substitute \(f^{-1}(x)\) into the original function \(f(x)\), get \(f(f^{-1}(x)) = \frac{f^{-1}(x)-1}{5} = \frac{5x+1-1}{5} = x\). This verifies the first equation.

Step 3: Verifying the second equation \(f^{-1}(f(x)) = x\)

To verify \(f^{-1}(f(x)) = x\), substitute \(f(x)\) into the inverse function \(f^{-1}(x)\), get \(f^{-1}(f(x)) = 5f(x) + 1 = 5(\frac{x-1}{5}) + 1 = x\). This verifies the second equation.

Key Concepts

Finding Inverse Functions

The journey to understand inverse functions in precalculus begins with a simple yet powerful idea: flipping the 'x' and 'y' in a function. To illustrate, let's use an exercise example where we have a function defined as f(x) = (x - 1)/5.

First, substitute 'y' for 'f(x)' to have y = (x - 1)/5. Next, exchange 'x' and 'y' to get x = (y - 1)/5. This swapping allows us to see what we must do to find our inverse. Then, solve for 'y' by performing algebraic manipulations, which will provide you with the inverse function. Remember, the inverse function will take an output of the original function and tell you the corresponding input. This process is like retracing your steps in a maze to get back to the start.

Verifying Inverse Functions

Your journey doesn't end once you've found a function that looks like an inverse; you must also check your work. Verification is a crucial step to ensure that the function you've found genuinely acts as an inverse. Here's how you do it: take the inverse function you derived and plug it into the original function. If the output simplifies down to 'x', you have verified the inverse correctly.

For our example, when we substitute f^{-1}(x) = 5x + 1 into the original function, and the result is 'x', it's like completing a magical spell where everything reverts back to its original form. The same applies when you put the original function inside the inverse function. This double-check confirms that both functions are true inverses of each other.

Inverse Function Notation

Now let's talk notation. The inverse of a function 'f' is denoted by f^{-1}. This doesn't mean 'f' to the power of negative one; it's just the symbol mathematicians agreed on to indicate an inverse function. Think of it as the 'undo' function for 'f'.

Using our example, if f(x) = (x - 1)/5, then the inverse, denoted as f^{-1}(x), does the opposite operation, which is 5x + 1 in this case. Remember, notation is a language of its own in math, and learning to read and write it correctly can make a world of difference in understanding and communicating mathematical concepts.

Algebraic Manipulation

Algebraic manipulation is the artist's brush in the world of mathematics; it's what you use to sculpt equations into the forms you need. When finding inverses, algebraic manipulation involves rearranging terms, factoring, or performing operations on both sides of an equation to isolate the desired variable.

In our given function, we multiplied both sides by 5 and then added 1 to isolate 'y'. Those simple moves are the essence of algebraic manipulation – transforming an equation step by step until it reveals the solution. The ability to manipulate equations is a fundamental skill that will pay off in numerous areas of mathematics and problem-solving.