Answer :
Answer: (c)
Step-by-step explanation:
Given
[tex]f(x)=\dfrac{1}{x-3}\\\\g(x)=\sqrt{x+5}[/tex]
Here, [tex]\sqrt{x+5}\ \text{is always greater than equal to 0}\\\Rightarrow x+5\geq 0\\\Rightarrow x\geq -5\quad \ldots(i)[/tex]
To get [tex]f\left(g(x)\right)[/tex], replace [tex]x[/tex] in [tex]f(x)[/tex] by [tex]g(x)\ \text{i.e. by}\ \sqrt{x+5}[/tex]
[tex]\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)[/tex]
Using [tex](i)[/tex] and [tex](ii)[/tex] it can be concluded that the domain of [tex]f\left(g(x)\right)[/tex] is all real numbers except 0.
Therefore, its domain is given by
[tex]x\in [-5,4)\cup (4,\infty)[/tex]
Option (c) is correct.
Answer:
its d
Step-by-step explanation:
i did the iready diagnostic test. :)