1. **State the problem:** We are given the second derivative of a function $f''(x) = x^{-3/2}$, with initial conditions $f'(4) = 2$ and $f(0) = 0$. We need to find the function $f(x)$. 2. **Recall the formulas:** To find $f(x)$, we integrate $f''(x)$ twice: $$f'(x) = \int f''(x) \, dx + C_1$$ $$f(x) = \int f'(x) \, dx + C_2$$ 3. **Integrate $f''(x)$ to find $f'(x)$:** $$f'(x) = \int x^{-3/2} \, dx + C_1$$ Recall that $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$ for $n \neq -1$. Here, $n = -\frac{3}{2}$, so: $$f'(x) = \frac{x^{-3/2 + 1}}{-3/2 + 1} + C_1 = \frac{x^{-1/2}}{-1/2} + C_1 = -2x^{-1/2} + C_1$$ 4. **Use initial condition $f'(4) = 2$ to find $C_1$:** $$2 = f'(4) = -2(4)^{-1/2} + C_1 = -2 \times \frac{1}{2} + C_1 = -1 + C_1$$ So, $$C_1 = 3$$ 5. **Integrate $f'(x)$ to find $f(x)$:** $$f(x) = \int (-2x^{-1/2} + 3) \, dx + C_2 = \int -2x^{-1/2} \, dx + \int 3 \, dx + C_2$$ Calculate each integral: $$\int x^{-1/2} \, dx = 2x^{1/2} + C$$ So, $$f(x) = -2 \times 2x^{1/2} + 3x + C_2 = -4x^{1/2} + 3x + C_2$$ 6. **Use initial condition $f(0) = 0$ to find $C_2$:** $$0 = f(0) = -4 \times 0^{1/2} + 3 \times 0 + C_2 = C_2$$ So, $$C_2 = 0$$ 7. **Final answer:** $$\boxed{f(x) = 3x - 4\sqrt{x}}$$ This function satisfies the given second derivative and initial conditions.