Find the inverse of the following one-to-one function. Give the domain and the range of \( f \) and of \( f^{-1} \), and then graph both \( f \) and \( f^{-1} \) on the same set of axes. \( f(x)=\frac{2}{x} \) \( f^{-1}(x)=\frac{2}{x} \) The domain of \( f \) is \( \square \) and the range of \( f \) is (Type your answers in interval notation.)

To find the domain and range of the function \( f(x) = \frac{2}{x} \), we start with the domain. Since \( x \) cannot be zero (as this would make the denominator undefined), the domain of \( f \) is \( (-\infty, 0) \cup (0, \infty) \). Now, let's determine the range. As \( x \) approaches zero from either side, \( f(x) \) approaches \( \infty \) or \( -\infty \), and as \( x \) approaches \( \pm\infty \), \( f(x) \) approaches zero but never actually reaches it. Therefore, the range of \( f \) is also \( (-\infty, 0) \cup (0, \infty) \). For the inverse function \( f^{-1}(x) = \frac{2}{x} \), we can see that it exhibits the same behavior as \( f(x) \). Thus, the domain of \( f^{-1} \) is \( (-\infty, 0) \cup (0, \infty) \) (since it can't be zero), and the range of \( f^{-1} \) is also \( (-\infty, 0) \cup (0, \infty) \). To graph both functions on the same axes, you can plot the hyperbola represented by \( f(x) = \frac{2}{x} \) — which hugs the axes, never crossing them— and its inverse, which is, coincidentally, the same function. Their symmetry can be illustrated beautifully by the line \( y = x \) as the reflection line. In conclusion: - Domain of \( f \): \( (-\infty, 0) \cup (0, \infty) \) - Range of \( f \): \( (-\infty, 0) \cup (0, \infty) \) - Domain of \( f^{-1} \): \( (-\infty, 0) \cup (0, \infty) \) - Range of \( f^{-1} \): \( (-\infty, 0) \cup (0, \infty) \) Now, grab your graphing materials and have fun plotting those curves!