College Algebra

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The vertical lines indicate the points at which the real line is divided into intervals.

Slide 42

E.g. 3—A Quadratic Inequality

We read from the table or the diagram that (x – 2)(x – 3) is negative on the interval (2, 3).

Slide 43

E.g. 3—A Quadratic Inequality

Thus, the solution of the inequality (x – 2)(x – 3) ≤ 0 is: {x | 2 ≤ x ≤ 3} = [2, 3]

We have included the endpoints 2 and 3 because we seek values of x such that the product is either less than or equal to zero.

Slide 44

E.g. 3—A Quadratic Inequality

The solution is illustrated here.

Slide 45

E.g. 4—Solving an Inequality

Solve the inequality

2x2 – x > 1

First, we move all the terms to the left-hand side 2x2 – x – 1 > 0

Factoring the left side of the inequality, we get (2x + 1)(x – 1) > 0

Slide 46

E.g. 4—Solving an Inequality

The factors of the left-hand side are 2x + 1 and x – 1.

These factors are zero when x is –1/2 and 1, respectively.

These numbers divide the real line into the intervals (–∞, –1/2), (–1/2, 1), (1, ∞)

Slide 47

E.g. 4—Solving an Inequality

We make the following diagram, using test points to determine the sign of each factor in each interval.

Slide 48

E.g. 4—Solving an Inequality

From the diagram, we see that (2x + 1)(x – 1) > 0

for x in the interval (–∞, –1/2) or (1, ∞).

So the solution set is the union of these two intervals: (–∞, –1/2)U(1, ∞)

Slide 49

E.g. 4—Solving an Inequality

The solution is illustrated here.

Slide 50

E.g. 5—Solving an Inequality with Repeated Factors

Solve the inequality

x(x – 1)2(x – 3) < 0

All nonzero terms are already on one side of the inequality.

Also, the nonzero side of the inequality is already factored.

So we begin by finding the intervals for this inequality.

Slide 51

E.g. 5—Solving an Inequality with Repeated Factors

The factors of the left-hand side are x, (x – 1)2, and x – 3.

These factors are zero when x = 0, 1, 3.