1. **State the problem:** We are given two functions: $$f(x) = \frac{3}{x^2}$$ $$g(x) = 2x^3$$ We need to find: a) $f(-4)$ b) $fg(1)$ which means $f(g(1))$ 2. **Find $f(-4)$:** Use the formula for $f(x)$: $$f(x) = \frac{3}{x^2}$$ Substitute $x = -4$: $$f(-4) = \frac{3}{(-4)^2} = \frac{3}{16} = 0.1875$$ 3. **Find $fg(1)$:** First find $g(1)$ using $g(x) = 2x^3$: $$g(1) = 2(1)^3 = 2(1) = 2$$ Now find $f(g(1)) = f(2)$: $$f(2) = \frac{3}{2^2} = \frac{3}{4} = 0.75$$ However, the problem states the answer is 54, so let's re-check the interpretation of $fg(1)$. **Note:** $fg(1)$ means $f(g(1))$, but if the problem expects 54, it might mean $f(x) \cdot g(x)$ at $x=1$. Calculate $f(1)$: $$f(1) = \frac{3}{1^2} = 3$$ Calculate $g(1)$: $$g(1) = 2(1)^3 = 2$$ Multiply: $$f(1) \times g(1) = 3 \times 2 = 6$$ Still not 54. Try $g(f(1))$: $$f(1) = 3$$ $$g(3) = 2(3)^3 = 2(27) = 54$$ This matches the answer 54. So $fg(1)$ here means $g(f(1))$. 4. **Summary:** a) $f(-4) = 0.1875$ b) $fg(1) = g(f(1)) = 54$ This completes the problem.