Question

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

Answer 1

fog means f(x) compsed with g(x) or f(g(x)

so put g(x) for the x in f(x)

f(g(x))=5/(g(x))
f(g(x))=5/(2x^2+5x)

domian is the numbers you can use
basically it is not in the domain if it makes the function undefined
some things not in domain are values that make the function divide by zero and/or take the square root of a negative number

so divide by 0
set denomenator to 0 and solve for restricted values
0=2x^2+5x
0=x(2x+5)
0=x or
0=2x+5
-5=2x
-2.5=x

the 2 values not in the domain of f o g are 0 and -2.5

Answer 2

two number not included in domain are x=0 and x=-5/2

Explanation:

Let f(x)= 5/x and g(x)=2x^2+5x

WE need to find out the excluded value in the domain of f o g

f o g is the composition of functions

f o g= f(g(x))

lets plug in g(x) in f(x)

`f(g(x))= f( 2x^2+5x)`

replace 2x^2+5x in f(x)

`f(g(x))= f(2x^2+5x)=(5)/(2x^2+5x)`

here we have 2x^2+5x in the denominator

to find the numbers that are in domain we set the denominator =0 and solve for x

Because when denominator becomes 0 the function is undefined

`2x^2+5x=0`

`x(2x+5)=0`

x=0 and 2x+5=0 (subtract 5 on both sides)

2x= -5 (divide by 2 on both sides)

x=-5/2