Problem 43 - 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 \(f^{-1}\) $$f(x)=(x-1)^{2}, x \leq 1$$

Step-by-Step Solution

Answer

The inverse of the function \(f(x) = (x - 1)^2\), where \(x \leq 1\), is \(f^{-1}(x) = -\sqrt{x} + 1\). The graph of both functions is a set of mirror images across the line \(y = x\). The domain and the range of \(f(x)\) are \([-∞, 1]\) and \([0, +∞)\) respectively, while for its inverse function \(f^{-1}(x)\), the domain and the range are \([0, +∞)\) and \([-∞, 1]\) respectively.

Step 1: Finding the Inverse Function

Replace \(f(x)\) with \(y\), we get \(y = (x - 1)^2\). To find \(f^{-1}(x)\), switch \(x\) and \(y\) to get \(x = (y - 1)^2\). Now, solve for \(y\) to get the inverse function. Taking the square root of both sides gives us two possible solutions \(y = \sqrt{x} + 1\) and \(y = -\sqrt{x} + 1\). Since \(x \leq 1\) for the original function \(f(x)\), the inverse function is \(f^{-1}(x) = -\sqrt{x} + 1\).

Step 2: Graphing the Function and Its Inverse

To graph \(f(x) = (x - 1)^2\) and its inverse \(f^{-1}(x) = -\sqrt{x} + 1\), remember that the graph of an inverse function is a reflection of the original function's graph over the line \(y = x\). Thus, the graph of \(f(x)\) is a parabola opening right with vertex at (1, 0), and the graph of \(f^{-1}(x)\) is a reflection of this graph over the line \(y = x\), which is a sideways parabola opening downward with vertex at (1, 0).

Step 3: Finding the Domain and Range

The domain of a function is the set of all allowable inputs (values of \(x\)) and the range is the set of all possible output values (values of \(f(x)\)). For the function \(f(x) = (x - 1)^2\), the domain is \([-∞, 1]\) and the range is \([0, +∞)\). Its inverse \(f^{-1}(x) = -\sqrt{x} + 1\), will have its domain and range as the range and domain of the original function respectively. Thus, the domain for \(f^{-1}\) is \([0, +∞)\) and the range is \([-∞, 1]\).