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anyanavicka [17]
3 years ago
11

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

Mathematics
2 answers:
Igoryamba3 years ago
5 0

Answer:

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

Step-by-step 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)=\frac{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

Yuri [45]3 years ago
5 0
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
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A coin bank that excepts only nickels and dimes contains $9.15. There are three more than twice as many nickels as there are dim
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<em><u>Solution:</u></em>

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