NO LINKS OR ELSE YOU'LL BE REPORTED! Only answer if you're very good at Math. Let f(x) = 1/x-3 and g(x) = √x + 5.What is the domain of ( f o g) (x)?

A: [-5,3]U[3,&]

B: [-5,-4]U[-4,&)

C:[-5,4]U[4,&]

D: [-5,-3]U [-3,&]

The & equals
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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 answerhope it's help

mx+89b(-98)nb

step-by-step explanation:

step-by-step explanation:

what grade level is this? need to know so i can look in my old textbooks!