1. **State the problem:** We are given two functions $f(x) = \frac{3 + x}{7}$ and $h(x) = 2x + 3$. We need to find: (a) $h(4)$ (b) $f(h(x))$ (c) $x$ such that $f(x) = h(x)$ (d) the inverse function $h^{-1}(x)$. 2. **Find (a) $h(4)$:** Substitute $x=4$ into $h(x)$: $$h(4) = 2(4) + 3 = 8 + 3 = 11$$ 3. **Find (b) $f(h(x))$:** This means substitute $h(x)$ into $f$: $$f(h(x)) = f(2x + 3) = \frac{3 + (2x + 3)}{7} = \frac{3 + 2x + 3}{7} = \frac{2x + 6}{7}$$ 4. **Find (c) $x$ such that $f(x) = h(x)$:** Set the two functions equal: $$\frac{3 + x}{7} = 2x + 3$$ Multiply both sides by 7: $$3 + x = 7(2x + 3)$$ $$3 + x = 14x + 21$$ Bring all terms to one side: $$3 + x - 14x - 21 = 0$$ $$-13x - 18 = 0$$ $$-13x = 18$$ $$x = \frac{18}{-13} = -\frac{18}{13}$$ 5. **Find (d) the inverse function $h^{-1}(x)$:** Start with $y = h(x) = 2x + 3$ Swap $x$ and $y$: $$x = 2y + 3$$ Solve for $y$: $$x - 3 = 2y$$ $$y = \frac{x - 3}{2}$$ So, $$h^{-1}(x) = \frac{x - 3}{2}$$