Problem 28 - Thomas Calculus

Question

Each of Exercises \(25-34\) gives a formula for a function \(y=f(x)\) . In each case, find \(f^{-1}(x)\) and identify the domain and range of \(f^{-1}\) . As a check, show that \(f\left(f^{-1}(x)\right)=f^{-1}(f(x))=x.\) $$f(x)=(1 / 2) x-7 / 2$$

Step-by-Step Solution

Answer

The inverse of the function is \(f^{-1}(x) = \frac{x+7}{2}\). Domain and range: all real numbers. Verification confirms the inverse is correct.

Step 1: Understanding the Given Function

We start with the given function \(f(x) = \frac{1}{2}x - \frac{7}{2}\). Our task is to find the inverse function \(f^{-1}(x)\). An inverse function essentially reverses the role of inputs and outputs, meaning we need to solve for \(x\) in terms of \(y\) and then express \(y\) in terms of \(x\).

Step 2: Finding the Inverse Function

To find \(f^{-1}(x)\), we first rewrite the function relationship \(y = \frac{1}{2}x - \frac{7}{2}\). To find the inverse, solve for \(x\) in terms of \(y\). Start by multiplying both sides by 2: \(2y = x - 7\). Then, add 7 to both sides: \(x = 2y + 7\). Finally, solve for \(y\): \(y = \frac{x + 7}{2}\). Thus, the inverse function is \(f^{-1}(x) = \frac{x + 7}{2}\).

Step 3: Identifying the Domain and Range

The domain of \(f(x)\) is all real numbers because there're no restrictions on \(x\) for \(f(x) = \frac{1}{2}x - \frac{7}{2}\). Similarly, for the inverse \(f^{-1}(x)\), since it is a linear function, the domain and range are also all real numbers, \((-finity, \infty)\).

Step 4: Verifying the Inverse Function

To verify, check the composition of functions. First, compute \(f\left(f^{-1}(x)\right)\). Replace \(x\) in \(f(x)\) with \(\frac{x + 7}{2}\): \(f\left(\frac{x + 7}{2}\right) = \frac{1}{2}\left(\frac{x + 7}{2}\right) - \frac{7}{2}\). Simplifying, it becomes \(\frac{1}{2} \cdot \frac{x + 7}{2} = \frac{x + 7}{4}\), then we subtract \(\frac{7}{2}\), resulting in \(x\). Similarly, compute \(f^{-1}(f(x))\): replace \(x\) in \(f^{-1}(x)\) with \(\frac{1}{2}x - \frac{7}{2}\), which simplifies to \(x\). Both results confirm \(f(f^{-1}(x)) = f^{-1}(f(x)) = x\).

Step 5: Conclusion

The inverse of the function \(f(x) = \frac{1}{2}x - \frac{7}{2}\) is \(f^{-1}(x) = \frac{x + 7}{2}\). The domain and range of \(f^{-1}(x)\) are both \((-finity, \infty)\), and we've verified the correctness of the inversion by composition.

Key Concepts

Domain and Range

The domain and range of a function are fundamental concepts in mathematics. For a function like our given linear function, the domain and range are typically all real numbers,

• The domain of a function is the set of all possible inputs (usually represented by the variable x) for which the function is defined. For a linear function like our example, it can take any real number as its input.
• The range is the set of all possible outputs (usually the variable y) the function can generate. Similarly, for a linear function, the range is all real numbers.

For the inverse function,

• The domain of the inverse function is the range of the original function; since the range of the original function is all real numbers, the domain for the inverse is also all real numbers.
• Conversely, the range of the inverse is the domain of the original function, which, as with the domain, results in all real numbers for the range as well.

Linear Function

A linear function is one of the simplest types of functions you'll encounter in algebra, familiar to many from the straight lines they produce on a graph. A linear function is generally written in the form:\[ y = mx + b \] where \(m\) represents the slope of the line and \(b\) represents the y-intercept, or where the line crosses the y-axis. In our example, \(f(x) = \frac{1}{2}x - \frac{7}{2}\),

• The slope \(m\) is \(\frac{1}{2}\) which indicates the line rises half a unit for every one unit it goes to the right.
• The y-intercept \(b\) is \(-\frac{7}{2}\), meaning the line crosses the y-axis at \(y=-3.5\).

Linear functions are significant due to their straightforward nature and prevalence in real-world scenarios, making them a cornerstone of algebra and calculus.

Composition of Functions

Composition of functions involves applying one function to the results of another, and it's a critical concept when working with inverse functions. When you compose two functions, \(f(g(x))\), you first apply \(g(x)\) and then apply \(f(x)\) to the result. The same steps are taken in reverse when doing \(g(f(x))\).

• When verifying inverse functions as in our exercise, you utilize this concept by showing \(f(f^{-1}(x)) = f^{-1}(f(x)) = x\).
• This means if you place the inverse function into the original function, it brings you back to where you started with the initial input \(x\).

This process confirms the validity of your inverse, affirming that it truly reverses the effects of the original function.