Question

Find the domain of (f(g(x))) when f(x)= square root of x-1 and g(x)= 1/4x​

Answer 1

`\large\boxed{Domain:\ x\in\left(0;\ (1)/(4)\right>\to0<x\leq(1)/(4)}`

Explanation:

`f(x)=√(x-1),\ g(x)=(1)/(4x)\\\\f\bigg(g(x)\bigg)-\text{instead of x in the function equation f(x) put}\ (1)/(4x):\\\\f\bigg(g(x)\bigg)=\sqrt{(1)/(4x)-1}=\sqrt{(1)/(4x)-(4x)/(4x)}=\sqrt{(1-4x)/(4x)}\\\\\text{The domain}\ D:\\\\(1-4x)/(4x)\geq0\ \wedge\ 4x\\eq0\\\\(1-4x)/(4x)\geq0\iff(1-4x)(4x)\geq0\\\\1-4x=0\to4x=1\to x=(1)/(4)\\\\4x=0\to x=0\\(look\ at\ the\ picture)\\\\(1)\qquadx\in\left<0,\ (1)/(4)\right>\\\\\\4x\\eq0\qquad\text{divide both sides by 4}\\\\(2)\qquad x\\eq0`

`\text{From}\ (1)\ \text{and}\ (2)\ \text{We have}\ x\in\left(0,\ (1)/(4)\right>`