Find \( g(0), g(-4), g(7) \), and \( g\left(\frac{3}{2}\right) \operatorname{lor} g(x)=\frac{x}{\sqrt{16-x^{2}}} \) Find \( g(0) \) Select the correct choice below and if nocessary, fill in the answor box to complete your choice A. \( g(0)=0 \) (Simplify your answer. Type an exacl onswer, using radicals as needed) B. \( \mathbf{g}(0) \) does not exist. C. \( g(0) \) is not a real number. Find \( g(-4) \). Select the correct choice bolow and, if necassary, fill in the answer box to complete your choice. A. \( 9(-4)= \) \( \square \) (Simplify your answer. Type an exact answer, using radicals as needed ) B. \( g(-4) \) does not exist. C. \( g(-4) \) is not a real number.

Let's evaluate the function \( g(x) = \frac{x}{\sqrt{16 - x^{2}}} \) at the specified points. ### 1. Find \( g(0) \) **Calculation:** \[ g(0) = \frac{0}{\sqrt{16 - (0)^2}} = \frac{0}{\sqrt{16}} = \frac{0}{4} = 0 \] **Correct Choice:** **A.** \( g(0) = 0 \) --- ### 2. Find \( g(-4) \) **Calculation:** \[ g(-4) = \frac{-4}{\sqrt{16 - (-4)^2}} = \frac{-4}{\sqrt{16 - 16}} = \frac{-4}{\sqrt{0}} \] Here, the denominator becomes \( \sqrt{0} = 0 \), which leads to division by zero. Division by zero is undefined in the real number system. **Correct Choice:** **B.** \( g(-4) \) does not exist. --- If you need evaluations for \( g(7) \) and \( g\left(\frac{3}{2}\right) \), feel free to ask!