Hi Paige;

1.) f(x)=(x^{2}-4x+3)/(x-1)

At first sight, we might guess that the domain is...

(-infinity,1)U(1, +infinity)

1 must be excluded because it would render (x-1) as zero.

However, the denominator can be factored with one of the parenthetical equations as (x-1)...

f(x)=[(x-3)(x-1)]/(x-1)

Let's verify (x-3)(x-1) with FOIL...

FIRST...x^{2}

OUTER...-1x

INNER...-3x

LAST...+3

x^{2}-x-3x+3

x^{2}-4x+3

f(x)=[(x-3)(x-1)]/(x-1)

Let's cancel...

f(x)=[(x-3)(x-1)]/(x-1)

f(x)=x-3

The domain is infinity.

2.) g(x)=(x^{2}-4)(x-3)/(x^{2}-x-6)

Let's factor (x^{2}-x-6)...

Is one of the parenthetical equations (x-3)?

(x-3)(x+2)

Let's FOIL...

FIRST...x^{2}

OUTER...2x

INNER...-3x

LAST...-6

x^{2}+2x-3x-6

x^{2}-x+6

(x^{2}-4)(x-3)/(x-3)(x+2)

Let's cancel where appropriate...

(x^{2}-4)(x-3)/(x-3)(x+2)

(x^{2}-4)/(x+2)

Initially, it would seem that the domain is infinity excluding -2 such that the numerator can render 0. However, it would seem that the current denominator can also be factored to include (x+2)...

(x+2)(x-2)/(x+2)

FIRST...x^{2}

OUTER...-2x

INNER...2x

LAST...-4

Let's cancel where appropriate...

(x+2)(x-2)/(x+2)

x-2

There are no limitations on the domain. It is from negative infinity to positive infinity.

3.) F(x)=(x^{3}-2x^{2})/(x-2)

Can the denominator be factored to include (x-2)?

x^{3}-2x^{2}

x(x^{2}-2x)

x(x-2)(x+2)

[x(x-2)(x+2)]/(x-2)

Let's cancel where appropriate...

[x(x-2)(x+2)]/(x-2)

There are no limitations on the domain.

4.) G(x)=√(x-1)

We already know that a negative number cannot be square rooted. Therefore, let's begin with the one number which would set the limit...

(x-1)?0

x?1

The domain is (1, +infinity)

5.) f(x)=3/√(x-4)

This equation cannot be factored.

We already know that the domain cannot include 4 because that would render the numerator as 0.

We already know that a negative number cannot be square rooted. Therefore, let's begin with the one number which would set the limit...

(x-4)?0

x?4

The domain is (4, +infinity).

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