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soldier1979 [14.2K]
2 years ago
12

Choose the domain for which each function is defined

Mathematics
1 answer:
klemol [59]2 years ago
6 0

Answer:

Part 1) f(x)=\frac{x+4}{x} -----> x\neq 0

Part 2) f(x)=\frac{x}{x+4} ----> x\neq -4

Part 3)  f(x)=x(x+4) ----> All real numbers

Part 4) f(x)=\frac{4}{x^2+8x+16} ----> x\neq -4

Step-by-step explanation:

we know that

The domain of a function is the set of all possible values of x

Part 1) we have

f(x)=\frac{x+4}{x}

we know that

In a quotient the denominator cannot be equal to zero

so

For the value of x=0 the function is not defined

therefore

The domain is

x\neq 0

Part 2) we have

f(x)=\frac{x}{x+4}

we know that

In a quotient the denominator cannot be equal to zero

so

For the value of x=-4 the function is not defined

therefore

The domain is

x\neq -4

Part 3) we have

f(x)=x(x+4)

Applying the distributive property

f*(x)=x^2+4x

This is a vertical parabola open upward

The function is defined by all the values of x

therefore

The domain is all real numbers

Part 4) we have

f(x)=\frac{4}{x^2+8x+16}

we know that

In a quotient the denominator cannot be equal to zero

so

Equate the denominator to zero

x^2+8x+16=0

Remember that

x^2+8x+16=(x+4)^2

(x+4)^2=0

The solution is x=-4

so

For the value of x=-4 the function is not defined

therefore

The domain is

x\neq -4

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Answer18:

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Answer19:

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Step-by-step explanation:

For question 18:

Given that vertices of a quadrilateral are A(-4,-1), B(-4,6), C(2,6) and D(2,-4)

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Given that vertices of a quadrilateral are A(-2,3), B(3,2), C(2,-1) and D(-3,0)

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