Problem 22 - Precalculus Essentials

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 $$f\left(f^{-1}(x)\right)=x \text { and } f^{-1}(f(x))=x$$ $$f(x)=\frac{2}{x}$$

Step-by-Step Solution

Answer

The inverse function is \(f^{-1}(x) = \frac{2}{x}\). The inverse function has been verified as both \(f(f^{-1}(x))=x\) and \(f^{-1}(f(x))=x\) are valid.

Step 1: Find the inverse function

To find the inverse of the function \(f(x)=\frac{2}{x}\), let’s first swap x and y in the equation. That gives: \(x = \frac{2}{y}\). To solve this for y, we can multiply both sides of the equation by y and then divide by x. That results in: \(y = \frac{2}{x}\). The inverse function \(f^{-1}(x)\) is hence \(f^{-1}(x)= \frac{2}{x}\).

Step 2: Verify the inverse function with \(f(f^{-1}(x))=x\)

Let us substitute \(f^{-1}(x)\) into \(f(x)\). This means, we replace x in the original equation of \(f(x)\) by \(\frac{2}{x}\). This gives: \(f(f^{-1}(x))=\frac{2}{\frac{2}{x}} = x\).

Step 3: Verify the inverse function with \(f^{-1}(f(x))=x\)

Now we substitute \(f(x)\) into \(f^{-1}(x)\). This means, we replace x in the equation of \(f^{-1}(x)\) by \(\frac{2}{x}\), which gives: \(f^{-1}(f(x))= \frac{2}{\frac{2}{x}} = x\).

Key Concepts

One-to-one Functions

A function is classified as one-to-one, or injective, when each element in the domain maps to a unique element in the range. This means every input has a distinct output. For the purpose of finding inverse functions, it’s crucial that the original function is one-to-one. Why? Because only then can the inverse map each output back to a unique input.

Consider the function given: \(f(x) = \frac{2}{x}\). In this case, every value of \(x\) has a corresponding and unique \(f(x)\), as no two different values of \(x\) will give the same output \(\frac{2}{x}\). This function is one-to-one, and therefore invertible.

To determine if a function is one-to-one, you can use the horizontal line test. Imagine drawing horizontal lines across the graph of the function. If no horizontal line intersects the graph more than once, the function is one-to-one. In our example, \(f(x) = \frac{2}{x}\) passes this test.

Algebraic Manipulation

Finding the inverse of a function often involves a technique called algebraic manipulation. It requires rearranging and solving the function's equation for the original variable after swapping dependent and independent variables.

For example, with the function \(f(x) = \frac{2}{x}\), swap \(x\) and \(y\): the equation becomes \(x = \frac{2}{y}\).

• Multiply both sides by \(y\) to eliminate the fraction: \(xy = 2\).
• Divide both sides by \(x\): \(y = \frac{2}{x}\).

Thus, the inverse function \(f^{-1}(x)\) is \(\frac{2}{x}\). This process showcases how algebra is used to rearrange equations and solve for one variable in terms of another, confirming the intuitive process of inversing a function.

Verification of Inverse Functions

Once an inverse function is found, it's crucial to verify its correctness. Verification checks whether the original function and its inverse indeed reverse each other's effect. There are two main checks for verification:

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

Let’s verify \(f(x) = \frac{2}{x}\) and \(f^{-1}(x) = \frac{2}{x}\):

- Substitute \(f^{-1}(x)\) into \(f(x)\):

• \(f\left(\frac{2}{x}\right) = \frac{2}{\frac{2}{x}} = x\).

- Substitute \(f(x)\) into \(f^{-1}(x)\):

• \(f^{-1}\left(\frac{2}{x}\right) = \frac{2}{\frac{2}{x}} = x\).

These computations show that applying \(f\) to \(f^{-1}\) or vice versa yields \(x\). Successfully verifying both equations confirms that the derived inverse is appropriate and that \(f^{-1}(x)\) accurately reverses the process initiated by \(f(x)\). This detailed verification guarantees the precision and correctness of the inverse function, crucial in mathematical problem solving.