Problem 44 - College Algebra 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 \(f^{-1}\) $$f(x)=(x-1)^{2}, x \geq 1$$

Step-by-Step Solution

Answer

The inverse function is \(f^{-1}(x) = \sqrt{x}+1\) with domain [0, \(∞\)) and range [1, \(∞\)). The function \(f(x) = (x-1)^2\) has domain [1, \(∞\)) and range [0, \(∞\)). Both the function and its inverse are graphed in a coordinate system with the function being a parabola opening upwards, and the inverse being a square-root graph shifted up one unit.

Step 1: Finding the inverse

To find the inverse function, let \(y = f(x) = (x-1)^{2}\) where \(x \geq 1\). Swap \(x\) and \(y\), so now it's \(x = (y-1)^2\). Solve for \(y\). Take the square root on both sides, \(y-1 = \sqrt{x}\), then isolate \(y\) by adding 1 to both sides, \(y = \sqrt{x} +1\). Since \(x \geq 1\), then the inverse function is \(f^{-1}(x) = \sqrt{x} + 1\) where \(x \geq 0\).

Step 2: Graphing the function and its inverse

To graph the function and its inverse, plot points for each and draw the graph. The function \(f(x) = (x-1)^2\) opens upwards as a parabola with vertex (1,0). The inverse function \(f^{-1}(x) = \sqrt{x}+1\) is a square-root function shifted up by 1 unit.

Step 3: Finding the domain and range

The domain and range of a function and its inverse are essentially swapped. The function \(f(x) = (x-1)^2\) has domain [1, \(∞\)) and range [0, \(∞\)). Similarly, the inverse \(f^{-1}(x) = \sqrt{x}+1\) has a domain [0, \(∞\)) and a range [1, \(∞\)).