1. **Problem statement:** Given functions $h(x) = x + 1$, $g(x) = x^2$, and $f(x) = x - 2$, find various compositions of these functions. 2. **Recall function composition:** For functions $f$ and $g$, the composition $f \circ g(x) = f(g(x))$ means you apply $g$ first, then $f$ to the result. 3. **Calculate each composition:** (i) $f \circ g(x) = f(g(x)) = f(x^2) = x^2 - 2$ (ii) $g \circ f(x) = g(f(x)) = g(x - 2) = (x - 2)^2 = x^2 - 4x + 4$ (iii) $g \circ h(x) = g(h(x)) = g(x + 1) = (x + 1)^2 = x^2 + 2x + 1$ (iv) $h \circ g \circ f(x) = h(g(f(x))) = h(g(x - 2)) = h((x - 2)^2) = (x - 2)^2 + 1 = x^2 - 4x + 4 + 1 = x^2 - 4x + 5$ (v) $g \circ h \circ f(x) = g(h(f(x))) = g(f(x) + 1) = g(x - 2 + 1) = g(x - 1) = (x - 1)^2 = x^2 - 2x + 1$ (vi) $g \circ f \circ g(x) = g(f(g(x))) = g(f(x^2)) = g(x^2 - 2) = (x^2 - 2)^2 = x^4 - 4x^2 + 4$ (vii) $g \circ g \circ h(x) = g(g(h(x))) = g(g(x + 1)) = g((x + 1)^2) = ((x + 1)^2)^2 = (x + 1)^4$ (viii) $h \circ f \circ f(x) = h(f(f(x))) = h(f(x - 2)) = h((x - 2) - 2) = h(x - 4) = (x - 4) + 1 = x - 3$ 4. **Part (b):** Given $p(x) = x^2 - 2x + 1$, express $p$ as a composition of $f$, $g$, and $h$. Note that $p(x) = (x - 1)^2 = g(x - 1) = g(h(x - 2))$ but $h(x) = x + 1$, so $x - 1 = f(x) + 1 = h(f(x))$. Therefore, $p(x) = g(h(f(x)))$. 5. **Summary for part (a):** (i) $f \circ g(x) = x^2 - 2$ (ii) $g \circ f(x) = x^2 - 4x + 4$ (iii) $g \circ h(x) = x^2 + 2x + 1$ (iv) $h \circ g \circ f(x) = x^2 - 4x + 5$ (v) $g \circ h \circ f(x) = x^2 - 2x + 1$ (vi) $g \circ f \circ g(x) = x^4 - 4x^2 + 4$ (vii) $g \circ g \circ h(x) = (x + 1)^4$ (viii) $h \circ f \circ f(x) = x - 3$ 6. **Summary for part (b):** $p(x) = g \circ h \circ f(x)$ 7. **Problem 16 (i):** Given $f(x) = 3x + 1$ and $g(x) = 5x + c$, find $c$ such that $f \circ g = g \circ f$. Calculate $f \circ g(x) = f(g(x)) = f(5x + c) = 3(5x + c) + 1 = 15x + 3c + 1$ Calculate $g \circ f(x) = g(f(x)) = g(3x + 1) = 5(3x + 1) + c = 15x + 5 + c$ Set equal: $15x + 3c + 1 = 15x + 5 + c$ Subtract $15x$ both sides: $$\cancel{15x} + 3c + 1 = \cancel{15x} + 5 + c$$ Simplify: $$3c + 1 = 5 + c$$ Subtract $c$ both sides: $$\cancel{3c} + 1 = 5 + \cancel{c} + 2c$$ Actually, better to write: $$3c + 1 = 5 + c$$ Subtract $c$ both sides: $$3c - c + 1 = 5$$ $$2c + 1 = 5$$ Subtract 1 both sides: $$2c = 4$$ Divide both sides by 2: $$\frac{2c}{2} = \frac{4}{2}$$ $$c = 2$$ 8. **Problem 16 (ii):** Test $c=2$ by showing $f(g(3)) = g(f(3))$. Calculate $g(3) = 5(3) + 2 = 15 + 2 = 17$ Calculate $f(g(3)) = f(17) = 3(17) + 1 = 51 + 1 = 52$ Calculate $f(3) = 3(3) + 1 = 9 + 1 = 10$ Calculate $g(f(3)) = g(10) = 5(10) + 2 = 50 + 2 = 52$ Since $f(g(3)) = g(f(3)) = 52$, the value $c=2$ is correct.