Answer:
Given
\begin{gathered}f(x)=\dfrac{1}{x-3}\\\\g(x)=\sqrt{x+5}\end{gathered}
f(x)=
x−3
1
g(x)=
x+5
Here, \begin{gathered}\sqrt{x+5}\ \text{is always greater than equal to 0}\\\Rightarrow x+5\geq 0\\\Rightarrow x\geq -5\quad \ldots(i)\end{gathered}
x+5
is always greater than equal to 0
⇒x+5≥0
⇒x≥−5…(i)
To get f\left(g(x)\right)f(g(x)) , replace xx in f(x)f(x) by g(x)\ \text{i.e. by}\ \sqrt{x+5}g(x) i.e. by
x+5
\begin{gathered}\Rightarrow f\left(g(x)\right)=\dfrac{1}{\sqrt{x+5}-3}\\\\\text{Denominator must not be equal to 0}\\\\\therefore \sqrt{x+5}-3\neq0\\\Rightarrow \sqrt{x+5}\neq 3\\\Rightarrow x+5\neq 9\\\Rightarrow x\neq 4\quad \ldots(ii)\end{gathered}
⇒f(g(x))=
x+5
−3
1
Denominator must not be equal to 0
∴
x+5
−3
=0
⇒
x+5
=3
⇒x+5
=9
⇒x
=4…(ii)
Using (i)(i) and (ii)(ii) it can be concluded that the domain of f\left(g(x)\right)f(g(x)) is all real numbers except 0.
Therefore, its domain is given by
x\in [-5,4)\cup (4,\infty)x∈[−5,4)∪(4,∞)
Option (c) is correct.
Step-by-step explanation:
that's so hard to answer.....hope it's help
