Problem 53 - Precalculus with Limits

Question

In Exercises \(49-62,\) (a) find the inverse function of \(f\) (b) graph both \(f\) and \(f^{-1}\) on the same set of coordinate axes, (c) describe the relationship between the graphs of \(f\) and \(f^{-1}\) , and (d) state the domain and range of \(f\) and \(f^{-1}\) . $$ f(x)=\sqrt{4-x^{2}}, \quad 0 \leq x \leq 2 $$

Step-by-Step Solution

Answer

The inverse function is \(f^{-1}(x) = \sqrt{4 - x^2}\). Both \(f(x)\) and \(f^{-1}(x)\) are plotted as semi-circles. They reflect about the line \(y = x\). The domain and range for both the original function and its inverse are \(0 \leq x \leq 2\) and \(0 \leq y \leq 2\), respectively.

Step 1: Find the Inverse Function

We start by setting \(y = f(x)\), yielding the equation \(y = \sqrt{4-x^2}\). To find the inverse, we switch \(y\) and \(x\), then solve it for \(y\). This gives us \(x = \sqrt{4 - y^2}\). Solving for \(y\) gives us \(y = \sqrt{4 - x^2}\) and \(y = -\sqrt{4 - x^2}\). Since the given domain is \(0 \leq x \leq 2\), the inverse function becomes \(f^{-1}(x) = \sqrt{4 - x^2}\).

Step 2: Graph

Next, we graph both functions. \(f(x)\) is a semi-circle above the \(x\)-axis with radius 2, while \(f^{-1}(x)\) is a semi-circle to the right of the \(y\)-axis with the same radius. These graphs are mirror images over the line \(y=x\).

Step 3: Describe the Relationship

The function \(f(x)\) and its inverse \(f^{-1}(x)\) are reflections of each other across the line \(y = x\). When a function has an inverse, this is always the relationship between their graphs.

Step 4: Determine the Domain and Range

The original function \(f(x)\) is defined for \(0 \leq x \leq 2\), with range \(0 \leq y \leq 2\). For the inverse function \(f^{-1}(x)\), the domain and range reverse, making the domain \(0 \leq x \leq 2\) and the range \(0 \leq y \leq 2\).