Problem 25

Question

Find the inverse of the given function by using the "undoing process," and then verify that \(\left(f \circ f^{-1}\right)(x)=x\) and \(\left(f^{-1} \circ f\right)(x)=x\). (Objective 4) $$f(x)=\frac{4}{5} x$$

Step-by-Step Solution

Verified
Answer
Inverse: \( f^{-1}(x) = \frac{5}{4}x \); verifications confirm \( (f \circ f^{-1})(x) = x \) and \( (f^{-1} \circ f)(x) = x \).
1Step 1: Express the Function with y
First, express the function using the variable \( y \). Since we have the function \( f(x) = \frac{4}{5}x \), we can write it as \( y = \frac{4}{5}x \).
2Step 2: Swap x and y
To find the inverse, swap \( x \) and \( y \) in the equation. This gives us \( x = \frac{4}{5}y \).
3Step 3: Solve for y
Next, solve the equation for \( y \). Multiply both sides of the equation by \( \frac{5}{4} \) to isolate \( y \). Thus, \( y = \frac{5}{4}x \). This is the inverse function, \( f^{-1}(x) = \frac{5}{4}x \).
4Step 4: Verify \( (f \circ f^{-1})(x) = x \)
To verify this, compose \( f \) with \( f^{-1} \) using the expression for \( f(x) = \frac{4}{5}x \) and \( f^{-1}(x) = \frac{5}{4}x \). So, \( (f \circ f^{-1})(x) = f(f^{-1}(x)) = f\left(\frac{5}{4}x\right) = \frac{4}{5} \cdot \frac{5}{4}x = x \).
5Step 5: Verify \( (f^{-1} \circ f)(x) = x \)
Similarly, compose \( f^{-1} \) with \( f \). Thus, \( (f^{-1} \circ f)(x) = f^{-1}(f(x)) = f^{-1}\left(\frac{4}{5}x\right) = \frac{5}{4} \cdot \frac{4}{5}x = x \).

Key Concepts

Understanding AlgebraFunction VerificationComposition of Functions
Understanding Algebra
Algebra is a key area of mathematics that uses symbols and letters to represent numbers and express mathematical relationships. In the context of finding inverse functions, algebraic manipulation is crucial.

For example, in our problem, we started with the function \( f(x) = \frac{4}{5}x \). The process of finding its inverse involved algebraic techniques, such as expressing the function with a different variable, like \( y = \frac{4}{5}x \), and then swapping \( x \) and \( y \) to discover the inverse equation. These steps are typical applications of algebra when working with functions.

Important algebraic concepts that often come into play include:
  • Substituting variables.
  • Isolating variables to solve equations.
  • Using inverse operations, like multiplying to reverse division.
By understanding and applying these algebraic principles, you can solve a variety of mathematical problems effectively.
Function Verification
Function verification involves confirming that an inverse function is indeed the true inverse of the original function. Specifically, it means checking that the composite of the function and its inverse return the original input value.

In mathematical terms, for a function \( f \) and its inverse \( f^{-1} \), the verification requires two key conditions:
  • \((f \circ f^{-1})(x) = x\) or \(f(f^{-1}(x)) = x\).
  • \((f^{-1} \circ f)(x) = x\) or \(f^{-1}(f(x)) = x\).

The problem demonstrated this by showing that applying the original function \( f(x) = \frac{4}{5}x \) and the inverse \( f^{-1}(x) = \frac{5}{4}x \) to each other simplified back to \( x \). This validation approach ensures that the derived inverse function works correctly and completely reverses the original operation.
Composition of Functions
The composition of functions is a crucial operation in mathematics that helps explore the relationship between two functions. It involves applying one function to the result of another function. You can think of it as a mathematical assembly line where one function processes an input and feeds the output to the next function.

With our exercise, we performed function composition to verify the inverse. We had two compositions:
  • The composition \((f \circ f^{-1})(x)\) meant applying \( f \) to the output of \( f^{-1} \).
  • The reverse composition \((f^{-1} \circ f)(x)\) meant applying \( f^{-1} \) to the output of \( f \).

The result in both cases was the return of the original input \( x \). This confirms that \( f^{-1} \) truly is the inverse of \( f \). Composition is a powerful tool to see how two functions interact and ensure mathematical processes are being correctly reversed or connected.