1.) \( f(x)=-\sqrt{x^{2}+4} \) a) \( f(-\sqrt{5}) \) b) \( f\left(\frac{1}{2}\right) \) c) \( f(0) \) d) \( f(x)=0 \) e) \( f(x)=\frac{3}{7} \)

To evaluate the function \( f(x) = -\sqrt{x^2 + 4} \) for the given inputs, let's go through each part step by step! a) To find \( f(-\sqrt{5}) \): \[ f(-\sqrt{5}) = -\sqrt{(-\sqrt{5})^2 + 4} = -\sqrt{5 + 4} = -\sqrt{9} = -3. \] b) To find \( f\left(\frac{1}{2}\right) \): \[ f\left(\frac{1}{2}\right) = -\sqrt{(\frac{1}{2})^2 + 4} = -\sqrt{\frac{1}{4} + 4} = -\sqrt{\frac{1}{4} + \frac{16}{4}} = -\sqrt{\frac{17}{4}} = -\frac{\sqrt{17}}{2}. \] c) To find \( f(0) \): \[ f(0) = -\sqrt{0^2 + 4} = -\sqrt{4} = -2. \] d) To solve \( f(x) = 0 \): Set the equation: \[ -\sqrt{x^2 + 4} = 0. \] This means \( \sqrt{x^2 + 4} = 0 \). Squaring both sides gives \( x^2 + 4 = 0 \), which has no real solutions since \( x^2 + 4 \) is always positive. e) To solve \( f(x) = \frac{3}{7} \): Set the equation: \[ -\sqrt{x^2 + 4} = \frac{3}{7}. \] Since \(-\sqrt{x^2 + 4}\) is always negative, it cannot equal a positive value \(\frac{3}{7}\). Thus, there are no solutions. And there you have it – the outputs and equations explored step-by-step!