Problem 41 - 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 \(f^{-1}\) $$f(x)=x^{2}-4, x \geq 0$$

Step-by-Step Solution

Answer

The inverse of the function \(f(x) = x^2 - 4, x \geq 0\) is \(f^{-1}(x) = \sqrt{x + 4}, x \geq -4\). The domain and range of \(f(x)\) are \([0, \infty)\) and \([-4, \infty)\) respectively, while those of \(f^{-1}(x)\) are \([-4, \infty)\) and \([0, \infty)\) respectively.

Step 1: Finding the Inverse

In order to find the inverse \(f^{-1}(x)\) of a function \(f(x)\), it is necessary to replace \(f(x)\) with \(y\), interchange \(x\) and \(y\) and then solve for \(y\). Replacing \(f(x)\) with \(y\), we get \(y = x^2 - 4\). Switching \(x\) and \(y\), we then get \(x = y^2 - 4\). Solving this equation for \(y\), we have \(y = \sqrt{x + 4}\). Since the original function \(f(x)\) was defined for \(x \geq 0\), the inverse function \(f^{-1}(x) = \sqrt{x + 4}\) is defined for \(x \geq -4\). So \(f^{-1}(x) = \sqrt{x + 4}, x \geq -4\).

Step 2: Graphing the Function and Its Inverse

The graph of the initial function \(f(x) = x^2 - 4\) is a parabola, opening upwards with the vertex at \((0, -4)\) shifted down 4 units from the graph of \(y = x^2\). The graph of its inverse \(f^{-1}(x) = \sqrt{x+4}\) is a half parabola that opens to the right. The point of symmetry between the original function and its inverse is located at the origin (\(0, 0)\).

Step 3: Determining the Domain and Range

The domain of \(f(x) = x^2 - 4\) can be set as \(x \geq 0\) whereas the range is restricted to all real numbers that are greater than or equal to -4, i.e., \(y \geq -4\). On the other hand, the domain of \(f^{-1}(x) = \sqrt{x+4}\) is \(x \geq -4\) because it cannot take any values less than -4, while its range will be all non-negative real numbers, i.e., \(y \geq 0\).