Problem 4 - Precalculus : Building Concepts and Connections

Question

Use the definition of \(f(x)\) as given by the following table. $$\begin{array}{|r|r|} \hline x & f(x) \\ \hline -2 & 5 \\ \hline -1 & 3 \\ \hline 1 & -2 \\ \hline 4 & -1 \\ \hline \end{array}$$ Find \(\left(f^{-1} \circ f\right)(4)\)

Step-by-Step Solution

Answer

The value of \(\left(f^{-1} \circ f\right)(4)\) is 4.

Step 1: Understand the Expression

The operation \(\left(f^{-1} \circ f\right)(4)\) represents the composition of function \(f\) with its inverse \(f^{-1}\). This composition, as per the properties of functions and their inverses, simplifies to the identity function which outputs the input value directly. To put it in simpler words, \(f^{-1}(f(x)) = x\), where \(x\) is the input. In our exercise, we need to find \(\left(f^{-1} \circ f\right)(4)\), which means we replace \(x\) with 4 in our earlier result.

Step 2: Substitute the Value

Substitute \(x\) with 4 in \(f^{-1}(f(x)) = x\), to get \(f^{-1}(f(4)) = 4.\)

Step 3: Evaluate the Function's Value

From the given table, the function value at \(x = 4\) is \(f(4) = -1\). We can substitute this value back into the composition relation, \(f^{-1}(f(4)) = 4\), to get \(f^{-1}(-1) = 4\). Therefore, -1 must be the inverse function value corresponding to \(x = 4\), since \(f^{-1}(f(4)) = 4\).

Key Concepts

Function Composition

Function composition is like putting together two functions to make a new one. Imagine a machine where you put something in, and it goes through two steps before coming out. For functions, we often write this as \(f(g(x))\), which means you apply function \(g\) first, and then function \(f\). When we see something like \(\left(f^{-1} \circ f\right)(x)\), it simply means we're applying two processes: first \(f\), and then its inverse \(f^{-1}\). This will output the original input \(x\). Think of it as undoing what you just did. If \(f\) adds 2, \(f^{-1}\) will subtract 2 to bring you back. These operations together form what we call the identity function. In our exercise, the composition \(\left(f^{-1} \circ f\right)(4)\) simplifies directly to the input value, which in this case, is 4.

Identity Function

The identity function is a special type of function that always gives you back exactly what you put in. It's like a mirror for numbers. Mathematically, it's written as \(I(x) = x\). In simpler terms, whatever number you feed into it will come out unchanged. When we deal with function inverses, like \(f\) and \(f^{-1}\), their composition creates the identity function. This means if you start with \(x\), transform via \(f\), and then reverse that with \(f^{-1}\), you wind up with your original \(x\). So in exercises involving \(\left(f^{-1} \circ f\right)\), no matter what specific \(f\) does, this composition will always loop back to give you \(x\). For our problem, after identifying the right \(f\) and \(f^{-1}\) values, it asserts that \(\left(f^{-1} \circ f\right)(4) = 4\). The identity function ensures you land where you began.

Evaluating Functions

Evaluating functions means finding out what the function returns given a specific input. It's like asking, "What does the function tell me if I give it this number?" Using a table or formula helps us do this accurately.In our exercise, we use the table provided to find out what \(f(4)\) is. By checking the table, we learn that \(f(4)\) equals -1. This step is key for complex operations like inverse compositions. After finding \(f(4) = -1\), we then use this value to solve for the inverse. We need to find where -1 appears in the results column to nail down which input \(f^{-1}\) maps it back to. In our table, it's 4, confirming that \(f^{-1}(-1) = 4\). Mastering evaluation both for direct and inverse helps us navigate tricky problems with ease.