1. The problem states that the function $h$ is defined by $h(x) = \frac{1}{x}$ for $x \neq 0$. We are asked to find $h^2(x)$, $h^3(x)$, $h^4(x)$ in simplest form, and then find $h^{22}(x)$ and $h^{33}(x)$. 2. Recall that $h^n(x)$ means applying the function $h$ $n$ times in succession. For example, $h^2(x) = h(h(x))$, $h^3(x) = h(h(h(x)))$, and so on. 3. Calculate $h^2(x)$: $$ h^2(x) = h(h(x)) = h\left(\frac{1}{x}\right) = \frac{1}{\frac{1}{x}} = x $$ So, $h^2(x) = x$. 4. Calculate $h^3(x)$: $$ h^3(x) = h(h^2(x)) = h(x) = \frac{1}{x} $$ So, $h^3(x) = \frac{1}{x}$. 5. Calculate $h^4(x)$: $$ h^4(x) = h(h^3(x)) = h\left(\frac{1}{x}\right) = x $$ So, $h^4(x) = x$. 6. We observe a pattern: for even powers, $h^{2k}(x) = x$, and for odd powers, $h^{2k+1}(x) = \frac{1}{x}$. 7. Calculate $h^{22}(x)$: Since 22 is even, $$ h^{22}(x) = x $$ 8. Calculate $h^{33}(x)$: Since 33 is odd, $$ h^{33}(x) = \frac{1}{x} $$ Final answers: - $h^2(x) = x$ - $h^3(x) = \frac{1}{x}$ - $h^4(x) = x$ - $h^{22}(x) = x$ - $h^{33}(x) = \frac{1}{x}$