1. **State the problem:** We are given two functions: $$f(x) = \sqrt{2x + 4}$$ $$g(x) = x^2 + 1$$ We need to find formulas for the compositions: i. $f(f(x))$ ii. $f(g(x))$ iii. $g(f(x))$ iv. $g(g(x))$ 2. **Recall the composition rule:** For two functions $f$ and $g$, the composition $f(g(x))$ means we substitute $g(x)$ into $f$ wherever $x$ appears. 3. **Calculate each composition:** **i. Calculate $f(f(x))$:** Start with $f(x) = \sqrt{2x + 4}$. Substitute $f(x)$ into $f$: $$f(f(x)) = f\left(\sqrt{2x + 4}\right) = \sqrt{2\left(\sqrt{2x + 4}\right) + 4}$$ Simplify inside the square root: $$= \sqrt{2\sqrt{2x + 4} + 4}$$ This is the simplified formula for $f(f(x))$. **ii. Calculate $f(g(x))$:** Substitute $g(x) = x^2 + 1$ into $f$: $$f(g(x)) = f(x^2 + 1) = \sqrt{2(x^2 + 1) + 4}$$ Simplify inside the square root: $$= \sqrt{2x^2 + 2 + 4} = \sqrt{2x^2 + 6}$$ **iii. Calculate $g(f(x))$:** Substitute $f(x) = \sqrt{2x + 4}$ into $g$: $$g(f(x)) = g\left(\sqrt{2x + 4}\right) = \left(\sqrt{2x + 4}\right)^2 + 1$$ Simplify the square: $$= 2x + 4 + 1 = 2x + 5$$ **iv. Calculate $g(g(x))$:** Substitute $g(x) = x^2 + 1$ into $g$: $$g(g(x)) = g(x^2 + 1) = (x^2 + 1)^2 + 1$$ Expand the square: $$= (x^2)^2 + 2 \cdot x^2 \cdot 1 + 1^2 + 1 = x^4 + 2x^2 + 1 + 1 = x^4 + 2x^2 + 2$$ 4. **Summary of results:** $$f(f(x)) = \sqrt{2\sqrt{2x + 4} + 4}$$ $$f(g(x)) = \sqrt{2x^2 + 6}$$ $$g(f(x)) = 2x + 5$$ $$g(g(x)) = x^4 + 2x^2 + 2$$ These formulas represent the compositions requested.