1. **State the problem:** Given functions $f(x) = 2x + 3$, $g(x) = x^2 + 1$, and $h(x) = x^3$, find the following: (i) $f \circ g$, $g \circ f$, $f^2$, $g^2$ and prove that $(f \circ g) \circ h = f \circ (g \circ h)$. 2. **Recall function composition and powers:** - Composition: $(f \circ g)(x) = f(g(x))$ means apply $g$ first, then $f$. - Function powers: $f^2(x) = f(f(x))$ means apply $f$ twice. 3. **Calculate $f \circ g$:** $$(f \circ g)(x) = f(g(x)) = f(x^2 + 1) = 2(x^2 + 1) + 3 = 2x^2 + 2 + 3 = 2x^2 + 5$$ 4. **Calculate $g \circ f$:** $$(g \circ f)(x) = g(f(x)) = g(2x + 3) = (2x + 3)^2 + 1 = (2x + 3)^2 + 1$$ Expand: $$(2x + 3)^2 = 4x^2 + 12x + 9$$ So, $$(g \circ f)(x) = 4x^2 + 12x + 9 + 1 = 4x^2 + 12x + 10$$ 5. **Calculate $f^2$:** $$(f^2)(x) = f(f(x)) = f(2x + 3) = 2(2x + 3) + 3 = 4x + 6 + 3 = 4x + 9$$ 6. **Calculate $g^2$:** $$(g^2)(x) = g(g(x)) = g(x^2 + 1) = (x^2 + 1)^2 + 1$$ Expand: $$(x^2 + 1)^2 = x^4 + 2x^2 + 1$$ So, $$(g^2)(x) = x^4 + 2x^2 + 1 + 1 = x^4 + 2x^2 + 2$$ 7. **Prove $(f \circ g) \circ h = f \circ (g \circ h)$:** Calculate left side: $$(f \circ g) \circ h = (f \circ g)(h(x)) = f(g(h(x)))$$ Calculate right side: $$f \circ (g \circ h) = f((g \circ h)(x)) = f(g(h(x)))$$ Since both sides equal $f(g(h(x)))$, the equality holds. **Final answers:** - $f \circ g = 2x^2 + 5$ - $g \circ f = 4x^2 + 12x + 10$ - $f^2 = 4x + 9$ - $g^2 = x^4 + 2x^2 + 2$ - $(f \circ g) \circ h = f \circ (g \circ h)$ is true by function composition associativity.