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Section 3
Domain

Section 12
Rational Inequalities

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The set of x coordinates in a relation (also known as the input or the independent variable). The domain of a function is the set of all inputs for the function. The domain of a function may be stated explicitly.

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f(x)=x+3/x-2
Set the denominator in x+3/ x-2 equal to 0 to find where the expression is undefined.
x-2=0
add 2 to both sides of the equation
x=2
The domain is all values of x that makes the expression defined.
Interval Notation would be (-infinity,2)Union(2,infinity)

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F(x)= x+1/2-x
When there is a denominator, we want to include only values of the input that do not force the denominator to be zero. So, we will set the denominator equal to 0 and solve for X.
2-x=0
-x=-2
divide both sides by -1 to get rid of the negative in front of the x
x=2
The domain is all values of X that make the expression defined
(-infinity, 2)Union( 2, infinity)

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f(x)= sqrt(8-x) / (x^2 - 2x - 8)
Remove the sqrt over (8-x) then set greater than or equal to zero
8-x ≥ 0
Solve for x
x ≤ 8
This shows where the expression is defined.
Next
Set the denominator (x^2 - 2x - 8) equal to 0
x^2 - 2x - 8 = 0
Find the factors whose product is -8 and sum -2 and set equal to 0
(x-4)=0
(x+2)=0
Solve both for x
x=4
x=-2
This shows where the the expression is undefined

Lastly put into interval notation
(-Infinity, -2) U (-2,4) U (4,8)

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A rational inequality is an inequality which contains a rational expression.

The x-values that make the numerator zero and the x-values that make the denominator zero (undefined) will determine the number line test values.

Also a rational expression changes its sign only at its zeros and its undefined values.

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Rational Inequalities #1 (Harold J. Silva-Ponte)

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Rational Inequalities #1 (Harold J. Silva-Ponte)

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(5x-1)/(x+2)>7
Subtract both side by 7 resulting in ((5x-1)/(x+2))-7>0
Multiply 7/1 by (x+2) resulting in (5x-1-7x-14)/(x+2)>0
Simplify the faction to end up with (-2x-15)/(x+2)>0
Due to the rule that the denominator can't be 0, the x in (x+2) is x=-2
Add 15 to both sides resulting in -2x=15
Divide by -2 to both sides resulting in x=-15/2
Plug-in -15/2 to the original equation and that results in x=7
Plug in other numbers into the original equation so you can make your intervals based on your results
Your intervals should result in (-15/2,-2)

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Rational Inequalities #2 (Harold J. Silva-Ponte)

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Rational Inequalities #2 (Harold J. Silva-Ponte)

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(x^2-10x-200)/(6x^2+x-2)<=70
Simplify the equation into ((x+10)(x-20))/((3x+2)(2x-1))<=70
Due to the rule that the denominator can't be 0, the x in (3x+2) is x=-2/3 and the x in (2x-1) is x=1/2
Subtract 10 to both sides resulting in x=60
Add 20 to both sides resulting in x=90
Plug-in both -10 and 20 to the original equation and they indeed are less than or equal to 70
Plug in other numbers into the original equation so you can make your intervals based on your results
Your intervals should result in (-infinity, -2/3) U (1/2, infinity)

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(x-8/x) ≤ 3-x
Move all of the expression to the left side and set less than or equal to 0
(x-8/x)-3 + x ≤ 0
combine to get
(x-8/x)(-3x/x)+x ≤ 0
since there is a common denominator place all the numerators together
((x-8-3x)/x) +x ≤0
subtract 3x from x inside the fraction
((-2x-8)/x) + x ≤0
now factor out 2
(2(-x-4)/x) + x ≤0
multiply out the single x by x/x to give us a fraction
(2(-x-4)/x) + (x times x/x) ≤0
((2(-x-4)/x) + x(x)) all over x ≤0
Simplify the numerator to get
((x-4)(x+2))/x≤0
x=0
x=4
x= -2
use each root to create test intervals and plug value into original equation and then view results
x<-2.      (true)
-2<x<0.  (false)
0<x<4.   (true)
x>4.       (false)
Write out interval notation using these results
(-infinity, -2) U (0,4)