Restrict the domain of \( f(x) \) so that \( f \) is one-to-one. Then find \( f^{-1}(x) \). \( f(x)=1-x^{2} \)

To restrict the domain of the function \( f(x) = 1 - x^2 \) so that it is one-to-one, we need to identify an interval where the function is either strictly increasing or strictly decreasing. The function \( f(x) = 1 - x^2 \) is a downward-opening parabola. To find where it is increasing or decreasing, we can compute its derivative: \[ f'(x) = -2x \] The derivative \( f'(x) = -2x \) is zero at \( x = 0 \). This means that the function changes from increasing to decreasing at this point. Specifically: - For \( x < 0 \), \( f'(x) > 0 \) (the function is increasing). - For \( x > 0 \), \( f'(x) < 0 \) (the function is decreasing). To make \( f(x) \) one-to-one, we can restrict the domain to either \( (-\infty, 0] \) or \( [0, \infty) \). We will choose the interval \( (-\infty, 0] \) for this example. Now, we will find the inverse function \( f^{-1}(x) \). 1. Start with the equation \( y = f(x) \): \[ y = 1 - x^2 \] 2. Solve for \( x \) in terms of \( y \): \[ y = 1 - x^2 \implies x^2 = 1 - y \implies x = \pm \sqrt{1 - y} \] Since we restricted the domain to \( (-\infty, 0] \), we take the negative root: \[ x = -\sqrt{1 - y} \] 3. Now, we express \( x \) in terms of \( y \): \[ f^{-1}(y) = -\sqrt{1 - y} \] 4. To express the inverse function in terms of \( x \), we replace \( y \) with \( x \): \[ f^{-1}(x) = -\sqrt{1 - x} \] Thus, the restricted domain of \( f(x) = 1 - x^2 \) is \( (-\infty, 0] \), and the inverse function is: \[ f^{-1}(x) = -\sqrt{1 - x} \] for \( x \leq 1 \).