Problem 20 - College Algebra

Question

The functions are all one-to-one. For each function, a. Find an equation for \(f^{-1}(x),\) the inverse function. b. Verify that your equation is correct by showing that $$f\left(f^{-1}(x)\right)=x \text { and } f^{-1}(f(x))=x$$ $$ f(x)=(x-1)^{3} $$

Step-by-Step Solution

Answer

The inverse function, \(f^{-1}(x)\), is \(\sqrt[3]{x} + 1.\) It is the correct inverse function as \(f\left(f^{-1}(x)\right)=x\) and \(f^{-1}\left(f(x)\right)=x\), as was required to show.

Step 1: Swap x and y

Start by swapping x and y in the given function to get \(x=(y-1)^{3}\). This makes the formula for the inverse function.

Step 2: Solve for y

The result of step 1 can be solved for y: we first take the cubic root of both sides, which gives us \(y-1 = \sqrt[3]{x}\). Then, add 1 to both sides of the equation to isolate y, which gives us \(y = \sqrt[3]{x} + 1\). Therefore, the inverse function, denoted by \(f^{-1}(x)\), is \(f^{-1}(x) = \sqrt[3]{x} + 1\)

Step 3: Verify that \(f\left(f^{-1}(x)\right)=x\)

Substitute \(f^{-1}(x)\) into \(f(x)\), to get \(f\left(f^{-1}(x)\right) = \left(\sqrt[3]{x} + 1 - 1\right)^{3}\). After evaluations and simplifications, we see that \(f\left(f^{-1}(x)\right) = x, as \(f^{-1}(x) + 1 - 1\) is equal to \(f^{-1}(x)\), and \(\left(f^{-1}(x)\right)^{3}\) is equal to x.

Step 4: Verify that \(f^{-1}\left(f(x)\right) = x\)

Now substitute \(f(x)\) into \(f^{-1}(x)\), which yields, \(f^{-1}(f(x)) = \sqrt[3]{(x-1)^{3}} + 1\). Again, after evaluations and simplifications, we see that \(f^{-1}(f(x)) = x, as \(\sqrt[3]{(x-1)^{3}}\) is equal to \(x-1\), so \(\sqrt[3]{(x-1)^{3}} + 1\) equals to \(x-1+1 = x\).

Key Concepts

One-to-One Functions

A one-to-one function, also known as an injective function, is a special type of function where each input (or x-value) has a unique output (or y-value), and vice versa. This means that no two different inputs can produce the same output.
In simple terms, for two values, say a and b, if \( f(a) = f(b) \), then it must be true that \( a = b \). Understanding one-to-one functions is important when dealing with inverse functions because only one-to-one functions have inverses that are also functions. In the case of our exercise, we have \( f(x) = (x-1)^3 \), a classic example of a one-to-one function. This cubic function is one-to-one because it steadily increases or decreases without "turning back" at any point.
To verify that a function is one-to-one, you can use the horizontal line test: if no horizontal line crosses the graph of the function more than once, the function is one-to-one.

Cubic Root

The cubic root, denoted as \( \sqrt[3]{x} \), is the inverse operation of cubing a number. This operation answers the question: "What number, when cubed, gives me x?"
In our exercise, the cubic root is crucial in finding the inverse function. Since we begin with \( x = (y-1)^3 \), we take the cubic root to eliminate the power of three, helping us solve for y. The cubic root has some helpful properties:

• Unlike square roots, cubic roots exist for all real numbers, including negative ones. This is because cubing a negative number still results in a negative number (for example, \( (-2)^3 = -8 \)).
• The cubic root is itself a one-to-one function, meaning it has an inverse.

By understanding how cubic roots work, we can efficiently solve for inverse functions involving cubics. For our problem, taking the cubic root of both sides of \( x = (y-1)^3 \) leads us to the equation \( y = \sqrt[3]{x} + 1 \).

Function Verification

Verifying functions is a vital step in confirming that two operations are indeed inverses of each other. For functions \( f \) and \( f^{-1} \), we must show that both replacements return the original input. This means:

• \( f(f^{-1}(x)) = x \)
• \( f^{-1}(f(x)) = x \)

Let's break this down. 1. **\( f(f^{-1}(x)) = x \):** Here, we substitute \( f^{-1}(x) = \sqrt[3]{x} + 1 \) into \( f(x) \). This becomes \( (\sqrt[3]{x} + 1 - 1)^3 \), which simplifies back to \( x \) because the expression returns to its initial state by cancelling the adjustments.2. **\( f^{-1}(f(x)) = x \):** This is the reverse, where \( f(x) = (x-1)^3 \) gets substituted into \( f^{-1}(x) \), giving us \( \sqrt[3]{(x-1)^3} + 1 \). With simplification, it also returns to \( x \) since \( \sqrt[3]{(x-1)^3} = x - 1 \), and adding 1 gets us back to \( x \). Successful verification of both equations confirms that \( f(x) \) and \( f^{-1}(x) \) are indeed inverse functions, ensuring completeness and accuracy of our math work.