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How to solve polynomial and rational inequalities by putting them in the form f(x) > 0 or f(x) < 0, finding the boundary points, and testing values in each interval. Two examples of polynomial and rational inequalities are provided and solved step by step.
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Definition 0.1 A polynomial inequality is an inequality that can be put in one of the forms:
f (x) < 0 , f (x) > 0 , f (x) โค 0 , or f (x) โฅ 0
where f is a polynomial function.
Note how the x-intercepts divide the polynomial into sections whose values are positive and whose values are negative. The blue graph stresses how these boundary points are only potential points where the points in the graph change sign.
Property 0.1 Solving a polynomial inequality.
f (x) > 0 or f (x) < 0 ,
where f is a polynomial function.
Example 0.1 f (x) = x^2 โ 4 x + 3 < 0 x^2 โ 4 x + 3 = (x โ 1)(x โ 3) = 0 so x = 1 or x = 3 This breaks up the number line into 3 segments:(โโ, 1)
โ (1, 3)
โ (3, โ) Pick a test value from each interval:{ 0 , 2 , 4 } f (0) = 0^2 โ 4(0) + 3 = 3 > 0 So for x values in (โโ, 1), f (x) is positive. Donโt include this interval. f (2) = 2^2 โ 4(2) + 3 = โ 1 < 0 So for x values in (1, 3), f (x) is negative. Include this interval. f (4) = 4^2 โ 4(4) + 3 = 3 > 0 So for x values in (3, โ), f (x) is positive. Donโt include this interval. The solution is all values in (1, 3) = {x| 1 < x < 3 }
Example 0.2 f (x) = x^3 + 7x^2 โ x โ 7 < 0 x^3 +7x^2 โxโ7 = x^2 (x+7)โ(x+7) = (x+7)(x^2 โ1) = (x+7)(xโ1)(x+1) = 0 So x + 7 = 0 or x โ 1 = 0 or x + 1 = 0 So x = โ 7 or x = 1 or x = โ 1 This breaks up the number line into 4 segments: (โโ, โ7)
โ (โ 7 , โ1)
โ (โ 1 , 1)
โ (1, โ) Pick a test value from each interval: {โ 8 , โ 2 , 0 , 2 } f (โ8) = (โ8)^3 + 7(โ8)^2 โ (โ8) โ 7 = โ 63 < 0 So for x values in (โโ, โ7), f (x) is negative. Include this interval. f (โ2) = (โ2)^3 + 7(โ2)^2 โ (โ2) โ 7 = 15 > 0 So for x values in (โ 7 , โ1), f (x) is positive. Donโt include this interval.
x+
So the x-intercept occurs at x = 5. f (x) is undefined when x โ 2 = 0 when x = 2. This breaks up the number line into 3 segments: (โโ, โ5)
โ (โ 5 , 2)
โ (2, โ) Pick a test value from each interval: {โ 6 , 0 , 3 }
positive.
include this interval.
this interval. The solution is (โโ, โ5)
โ (2, โ) = {x|x < โ 5 or x > 2 }
What do we do when we do not have a 0 on one side of the equation? Put one there.
x
x
x
2(xโ1)
xโ2(xโ1)
xโ 2 x+
โx+
โx + 2 = 0 when x = 2 So the x-intercept occurs when x = 2. f (x) is undefined when x โ 1 = 0 when x = 1. This breaks up the number line into 3 segments: (โโ, 1)
โ (1, 2)
โ (2, โ) Pick a test value from each interval: { 0 , 32 , 3 }
include this interval.
f (^32 ) =
โ 32 + 3 2 โ^1
this interval.
include this interval. The solution is (1, 2) = {x| 1 < x < 2 }
