Problem 39 - Precalculus Essentials

Question

a. Find an equation for \(f^{-1}(x)\) b. Graph \(f\) and \(f^{-1}\) in the same rectangular coordinate system. c. Use interval notation to give the domain and the range of \(f\) and \(f^{-1}\) $$f(x)=2 x-1$$

Step-by-Step Solution

Answer

a. \(f^{-1}(x) = (x+1)/2\)\nb. Refer to above content for the graphical representation.\nc. The domain and range for both \(f\) and \(f^{-1}\) are \((-\infty,+\infty)\).

Step 1: Find the Inverse Function

To find the inverse of a function, one way is to swap \(x\) and \(y\), and then solve for \(y\). So for the function \(f(x)=2 x-1\), first change it to \(y = 2x - 1\). Swapping \(x\) and \(y\), we get \(x = 2y - 1\). Solving for \(y\), we find \(f^{-1}(x) = (x+1)/2\).

Step 2: Graph the Functions

To graph \(f\) and \(f^{-1}\) on the same coordinate system, plot the original function \(f(x) = 2x - 1\), which is a straight line with a slope of 2 and a y-intercept of -1. Then plot its inverse \(f^{-1}(x) = (x+1)/2\), which is also a straight line, but with a slope of 1/2 and a y-intercept of 1/2.

Step 3: Determine the Domains and Ranges

The domain of a function is the set of all real values of \(x\) that make the function defined. The range is the set of all the corresponding output values. For \(f(x)=2x-1\), the function is continuous and defined for all real values of \(x\), so the domain is \(-\infty,+\infty\) or \((-\infty,+\infty)\) in interval notation. The corresponding range of \(f(x)=2x-1\), is also \(-\infty,+\infty\) or \((-\infty,+\infty)\) in interval notation as the line can extend infinitely in both the positive and negative y direction. For \(f^{-1}(x)\), since it is the inverse of \(f\), its domain and range is the same as the range and domain of \(f\), and thus remains the same \(-\infty,+\infty\) or \((-\infty,+\infty)\).