1. **State the problem:** We are given two functions: $$f(x) = \frac{1}{\sqrt{x-1}}, \quad g(x) = (x^2 + 1)^2$$ We need to find the compositions: $$(f \circ g)(x) = f(g(x)) \quad \text{and} \quad (g \circ f)(x) = g(f(x))$$ 2. **Recall the composition formula:** For two functions $f$ and $g$, the composition $f \circ g$ means applying $g$ first, then $f$ to the result: $$ (f \circ g)(x) = f(g(x)) $$ Similarly, $$ (g \circ f)(x) = g(f(x)) $$ 3. **Find $(f \circ g)(x)$:** Substitute $g(x)$ into $f$: $$ (f \circ g)(x) = f\big((x^2 + 1)^2\big) = \frac{1}{\sqrt{(x^2 + 1)^2 - 1}} $$ Simplify inside the square root: $$ (x^2 + 1)^2 - 1 = (x^2 + 1)^2 - 1^2 = \big((x^2 + 1) - 1\big)\big((x^2 + 1) + 1\big) $$ $$ = (x^2)(x^2 + 2) = x^2(x^2 + 2) $$ So, $$ (f \circ g)(x) = \frac{1}{\sqrt{x^2(x^2 + 2)}} $$ Since $x^2 \geq 0$, we can write: $$ \sqrt{x^2(x^2 + 2)} = \sqrt{x^2} \sqrt{x^2 + 2} = |x| \sqrt{x^2 + 2} $$ Therefore, $$ (f \circ g)(x) = \frac{1}{|x| \sqrt{x^2 + 2}} $$ 4. **Find $(g \circ f)(x)$:** Substitute $f(x)$ into $g$: $$ (g \circ f)(x) = g\left(\frac{1}{\sqrt{x-1}}\right) = \left(\left(\frac{1}{\sqrt{x-1}}\right)^2 + 1\right)^2 $$ Simplify inside the parentheses: $$ \left(\frac{1}{x-1} + 1\right)^2 = \left(\frac{1}{x-1} + \frac{x-1}{x-1}\right)^2 = \left(\frac{1 + (x-1)}{x-1}\right)^2 = \left(\frac{x}{x-1}\right)^2 $$ So, $$ (g \circ f)(x) = \left(\frac{x}{x-1}\right)^2 = \frac{x^2}{(x-1)^2} $$ 5. **Domain considerations:** - For $f(x) = \frac{1}{\sqrt{x-1}}$, the domain is $x > 1$ because the expression under the square root must be positive. - For $g(x)$, the domain is all real numbers. - For $(f \circ g)(x)$, since $g(x) = (x^2 + 1)^2 \geq 0$, and $f$ requires input $>1$, check if $g(x) > 1$: $$ (x^2 + 1)^2 > 1 \implies x^2 + 1 > 1 \implies x^2 > 0 $$ which is true for all $x \neq 0$. At $x=0$, $g(0) = 1$, so $f(g(0))$ is undefined. - For $(g \circ f)(x)$, since $f(x)$ requires $x > 1$, the domain is $x > 1$. **Final answers:** $$ (f \circ g)(x) = \frac{1}{|x| \sqrt{x^2 + 2}} $$ $$ (g \circ f)(x) = \frac{x^2}{(x-1)^2} $$