Question

Let f(x) = 5/x and g(x)=2x2+5x. What two numbers are not in the domain of f o g​

Answer 1

`(f \circ g)(x)=(5)/(2x^2+5x)\\\\2x^2+5x\\ot =0\\x(2x+5)\\ot=0\\x\\ot =0 \wedge x\\ot =-(5)/(2)`

Answer 2

0 and -5/2

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