Answer:
0 and -5/2
Step-by-step explanation:
g is the first function we consider because that is the function we are first plugging in values into since the order is f o g and not g o f.
g has domain all real numbers meaning you can plug in any number into g and get a number back
So now let's look at plugging in g(x) into f(x)
that is f(g(x))=f(2x^2+5x)=5/(2x^2+5x)
Here you are dividing by a variable
You have to watch out dividing by 0
The variable, 2x^2+5x, is 0 when....
2x^2+5x=0
x(2x+5)=0
x=0 or x=-5/2
So The domain is all real numbers except x=0 or x=-5/2
[tex](f \circ g)(x)=\dfrac{5}{2x^2+5x}\\\\2x^2+5x\not =0\\x(2x+5)\not=0\\x\not =0 \wedge x\not =-\dfrac{5}{2}[/tex]
