Help me with this Suppose that \[ f(x)=\frac{1}{x} \text { and } g(x)=2 x^{3}-6 x \] For each function $h$ given below, find a formula for $h(x)$ and the domain of $h$. Enter the domains using interval notation a. $h(x)=(f \circ g)(x)=$ Domain: b. $h(x)=(g \circ f)(x)=$ Domain: c. $h(x)=(f \circ f)(x)=$ Domain: d. $h(x)=(g \circ g)(\theta)=$ Domain:

Step 1 ：\(h(x)=(f \circ g)(x)=f(g(x))=\frac{1}{g(x)}=\frac{1}{2x^3-6x}\)

Step 2 ：The domain of \(h(x)\) is the set of all real numbers \(x\) such that \(2x^3-6x\neq0\). This simplifies to \(x(x^2-3)\neq0\), so \(x\neq0\) and \(x\neq\pm\sqrt{3}\). Therefore, the domain of \(h(x)\) is \((-\infty,-\sqrt{3})\cup(-\sqrt{3},0)\cup(0,\sqrt{3})\cup(\sqrt{3},\infty)\).

Step 3 ：\(h(x)=(g \circ f)(x)=g(f(x))=2(f(x))^3-6f(x)=2\left(\frac{1}{x}\right)^3-6\left(\frac{1}{x}\right)=\frac{2}{x^3}-\frac{6}{x}\)

Step 4 ：The domain of \(h(x)\) is the set of all real numbers \(x\) such that \(x\neq0\). Therefore, the domain of \(h(x)\) is \((-\infty,0)\cup(0,\infty)\).

Step 5 ：\(h(x)=(f \circ f)(x)=f(f(x))=\frac{1}{f(x)}=\frac{1}{\frac{1}{x}}=x\)

Step 6 ：The domain of \(h(x)\) is the set of all real numbers \(x\) such that \(x\neq0\). Therefore, the domain of \(h(x)\) is \((-\infty,0)\cup(0,\infty)\).

Step 7 ：\(h(\theta)=(g \circ g)(\theta)=g(g(\theta))=2(g(\theta))^3-6g(\theta)=2\left(2\theta^3-6\theta\right)^3-6\left(2\theta^3-6\theta\right)\)

Step 8 ：The domain of \(h(\theta)\) is the set of all real numbers \(\theta\) such that \(2\theta^3-6\theta\neq0\). This simplifies to \(\theta(\theta^2-3)\neq0\), so \(\theta\neq0\) and \(\theta\neq\pm\sqrt{3}\). Therefore, the domain of \(h(\theta)\) is \((-\infty,-\sqrt{3})\cup(-\sqrt{3},0)\cup(0,\sqrt{3})\cup(\sqrt{3},\infty)\).