Question

Given Functions g(x) = 1/√x and f(x) = x^2 -4, state the domains of the following functions using interval notation.domain of g(x)/f(x):domain of g(f(x)):domain of f(g(x)):

Answer 1

We have the following functions

`g\mleft(x\mright)=\frac{1}{\sqrt[]{x}},f(x)=x^2-4`

Let's determine the domain of the following:

1. domain of g(x)/f(x):

`(g(x))/(f(x))=\frac{\frac{1}{\sqrt[]{x}}}{x^2-4}`

Let's simplify the above

`\begin{gathered} =\frac{1}{\sqrt[]{x}\cdot(x^2-4)} \\ =\frac{1}{\sqrt[]{x}}\cdot(1)/(x^2-4) \end{gathered}`

The above it's like having to functions, hence two domains:

`\begin{gathered} domain\mleft((1)/(√(x))\mright)\colon\mleft(0,\: \infty\: \mright) \\ domain\mleft((1)/(x^2-4)\mright)\colon\mleft(-\infty\: ,\: -2\mright)\cup\mleft(-2,\: 2\mright)\cup\mleft(2,\: \infty\: \mright) \end{gathered}`

Therefore, the domain of g/f (x) is:

`\mathrm{Domain\: of\: }\: (1)/(√(x)\left(x^2-4\right))\: \colon\mleft(0,\: 2\mright)\cup\mleft(2,\: \infty\: \mright)`

2. domain of g(f(x)):

`g(f(x))=g(x^2-4)=\frac{1}{\sqrt[]{x^2-4}}`

Since, x^2-4 > 0, therefore domain of g(f(x)) is the following:

`\mathrm{Domain\: of\: }\: (1)/(√(x^2-4))\: \colon\mleft(-\infty\: ,\: -2\mright)\cup\mleft(2,\: \infty\: \mright)`

3. domain of f(g(x)):​

`f(g(x))=f(\frac{1}{\sqrt[]{x}})=(\frac{1}{\sqrt[]{x}})^2-4`

Since, x>0, Therefore, the domain of f(g(x)) is:

`domain(\frac{1}{\sqrt[]{x}}^2-4)\colon(0,\infty)`