Find \( g(0), g(-4), g(7) \), and \( g\left(\frac{3}{2}\right) \) for \( g(x)=\frac{x}{\sqrt{16-x^{2}}} \). Find \( g(0) \). Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. \( g(0)=\square \) \( ( \) Simplify your answer. Type an exact answer, using radicals as needed.) B(0) does not exist. C. \( g(0) \) is not a real number.

To find \( g(0) \): Substituting \( x = 0 \) into the function: \[ g(0) = \frac{0}{\sqrt{16 - 0^2}} = \frac{0}{\sqrt{16}} = \frac{0}{4} = 0. \] So, the correct choice is: A. \( g(0) = 0 \). Now, let's evaluate \( g(-4) \): Substituting \( x = -4 \): \[ g(-4) = \frac{-4}{\sqrt{16 - (-4)^2}} = \frac{-4}{\sqrt{16 - 16}} = \frac{-4}{\sqrt{0}}. \] Since the denominator is zero, \( g(-4) \) does not exist. Next, let's find \( g(7) \): Substituting \( x = 7 \): \[ g(7) = \frac{7}{\sqrt{16 - 7^2}} = \frac{7}{\sqrt{16 - 49}} = \frac{7}{\sqrt{-33}}. \] Since the result involves a square root of a negative number, \( g(7) \) is not a real number. Lastly, let's calculate \( g\left(\frac{3}{2}\right) \): Substituting \( x = \frac{3}{2} \): \[ g\left(\frac{3}{2}\right) = \frac{\frac{3}{2}}{\sqrt{16 - \left(\frac{3}{2}\right)^2}} = \frac{\frac{3}{2}}{\sqrt{16 - \frac{9}{4}}} = \frac{\frac{3}{2}}{\sqrt{\frac{64}{4} - \frac{9}{4}}} = \frac{\frac{3}{2}}{\sqrt{\frac{55}{4}}} = \frac{\frac{3}{2}}{\frac{\sqrt{55}}{2}} = \frac{3}{\sqrt{55}}. \] In conclusion: - \( g(0) = 0 \) - \( g(-4) \) does not exist. - \( g(7) \) is not a real number. - \( g\left(\frac{3}{2}\right) = \frac{3}{\sqrt{55}} \).