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Expert-verified Found in: Page 241 ### Precalculus Enhanced with Graphing Utilities

Book edition 6th
Author(s) Sullivan
Pages 1200 pages
ISBN 9780321795465 # Solve the inequality algebraically$\mathbf{\left(}\mathbit{x}\mathbf{-}\mathbf{1}\mathbf{\right)}\mathbf{\left(}\mathbit{x}\mathbf{-}\mathbf{2}\mathbf{\right)}\mathbf{\left(}\mathbit{x}\mathbf{-}\mathbf{3}\mathbf{\right)}\mathbf{\le }\mathbf{0}$

Required solution set is

$\left(-\infty ,1\right]\cup \left[2,3\right]$

See the step by step solution

## Step 1.Given information

we have a given inequality

$\left(x-1\right)\left(x-2\right)\left(x-3\right)\le 0$

## Step 2.Finding the zeros

Zeroes of inequality

$f\left(x\right)=\left(x-1\right)\left(x-2\right)\left(x-3\right)\le 0$

are,

$x=1\phantom{\rule{0ex}{0ex}}x=2\phantom{\rule{0ex}{0ex}}x=3$

## Step 3.Divide the real number line into 4 intervals

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

$\left(-\infty ,1\right),\left(1,2\right),\left(2,3\right),\left(3,\infty \right)$

## 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 $\mathbit{f}\mathbf{\left(}\mathbit{x}\mathbf{\right)}\mathbf{=}\mathbf{\left(}\mathbit{x}\mathbf{-}\mathbf{1}\mathbf{\right)}\mathbf{\left(}\mathbit{x}\mathbf{-}\mathbf{2}\mathbf{\right)}\mathbf{\left(}\mathbit{x}\mathbf{-}\mathbf{3}\mathbf{\right)}\mathbf{=}\mathbf{0}$

is positive or negative.

In the interval $\left(-\infty ,1\right)$we chose 0, where f is negative

In the interval $\left(1,2\right)$

we chose 1.5 , where f is positive.

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

In the interval $\left(3,\infty \right)$

we chose 4 , where f is positive.

We know that our required inequality is$f\left(x\right)\le 0$

Here the inequality is strict $\left(\ge or\le \right)$ so we have to include the solutions of $f\left(x\right)=0$

in the solution set.

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

So required solution set is $\left(-\infty ,1\right]\cup \left[2,3\right]$ ### Want to see more solutions like these? 