1. **State the problem:** We are given two functions: $$f(x) = \frac{1}{2}x + 5$$ $$g(x) = x^2$$ We need to evaluate: (i) $g(3) + g(-3)$ (ii) $f^{-1}(6)$ (the inverse of $f$ evaluated at 6) (iii) $fg(2)$ (which means $f(g(2))$) 2. **Evaluate (i) $g(3) + g(-3)$:** Since $g(x) = x^2$, we have: $$g(3) = 3^2 = 9$$ $$g(-3) = (-3)^2 = 9$$ Adding these: $$g(3) + g(-3) = 9 + 9 = 18$$ 3. **Find $f^{-1}(6)$:** The inverse function $f^{-1}$ satisfies $f(f^{-1}(x)) = x$. Start with: $$y = f(x) = \frac{1}{2}x + 5$$ To find $f^{-1}$, solve for $x$ in terms of $y$: $$y = \frac{1}{2}x + 5$$ Subtract 5 from both sides: $$y - 5 = \frac{1}{2}x$$ Multiply both sides by 2: $$2(y - 5) = \cancel{2} \times \frac{1}{2}x = x$$ So, $$f^{-1}(y) = 2(y - 5)$$ Now evaluate at $y=6$: $$f^{-1}(6) = 2(6 - 5) = 2(1) = 2$$ 4. **Evaluate $fg(2)$:** This means $f(g(2))$. First find $g(2)$: $$g(2) = 2^2 = 4$$ Now find $f(4)$: $$f(4) = \frac{1}{2} \times 4 + 5 = 2 + 5 = 7$$ **Final answers:** (i) $18$ (ii) $2$ (iii) $7$