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Q. 25

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Precalculus Enhanced with Graphing Utilities
Found in: Page 241
Precalculus Enhanced with Graphing Utilities

Precalculus Enhanced with Graphing Utilities

Book edition 6th
Author(s) Sullivan
Pages 1200 pages
ISBN 9780321795465

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Short Answer

Solve the inequality algebraically

(x-1)(x-2)(x-3)0

Required solution set is

(-,1][2,3]

See the step by step solution

Step by Step Solution

Step 1.Given information 

we have a given inequality

(x-1)(x-2)(x-3)0

Step 2.Finding the zeros 

Zeroes of inequality

f(x)=(x-1)(x-2)(x-3)0

are,

x=1x=2x=3

Step 3.Divide the real number line into 4 intervals 

Now we use the zeros to separate the real number line into intervals.

(-,1),(1,2),(2,3),(3,)

Step 4.Selecting a test number in each interval 

Now we select a test number in each interval found in Step 3 and evaluate at each number to determine if f(x)=(x-1)(x-2)(x-3)=0

is positive or negative.

In the interval (-,1)we chose 0, where f is negative

In the interval (1,2)

we chose 1.5 , where f is positive.

In the interval (2,3) we chose 2.5 , where f is negative.

In the interval (3,)

we chose 4 , where f is positive.

We know that our required inequality isf(x)0

Here the inequality is strict (or ) so we have to include the solutions of f(x)=0

in the solution set.

So we want the interval where f is negative or equals to 0.

So required solution set is (-,1] [2,3]

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