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Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

Solutions

Chapter A, Problem 22E
To determine

To solve the below inequality in terms of intervals and illustrate the solution set on the real number line -

  (x+1)(x2)(x+3)0

Expert Solution

Answer to Problem 22E

The solution set of the inequality is 3x1 or x2 and the solution set on a real number line -

  Single Variable Calculus: Concepts and Contexts, Enhanced Edition, Chapter A, Problem 22E , additional homework tip  1

Explanation of Solution

Given: Inequality: (x+1)(x2)(x+3)0

Formula Used:

An inequality compares two values, showing if one is less than, greater than, or simply not equal to another value.

Real number line is the line whose points are the real numbers. 

Calculation:

Given: Inequality equation is (x+1)(x2)(x+3)0

Simplifying the above inequality, we have:

  (x+1)(x2)(x+3)0

To solve the above inequalities, we need to find the different intervals for which the inequality gives a value less than or equal to 0 .

When x<3 :

  (x+1) is negative, (x2) is negative and (x+3) is negative

Thus, (x+1)(x2)(x+3)0

So, x<3 isnot one of the solutions.

When x=3 :

  (x+1) is negative, (x2) is negative and (x+3) is Zero

Thus, (x+1)(x2)(x+3)=0

So, x=3 is one of the solutions.

When 3<x<1 :

  (x+1) is negative, (x2) is negative and (x+3) is positive

Thus, (x+1)(x2)(x+3)>0

So, 3<x<1 is one of the solutions.

When x=1 :

  (x+1) is Zero, (x2) is negative and (x+3) is positive

Thus, (x+1)(x2)(x+3)=0

So, x=1 is one of the solutions.

When 1<x<2 :

  (x+1) is positive, (x2) is negative and (x+3) is positive

Thus, (x+1)(x2)(x+3)<0

So, 1<x<2 is not one of the solutions.

When x=2 :

  (x+1) is positive, (x2) is zero and (x+3) is positive

Thus, (x+1)(x2)(x+3)=0

So, x=2 is one of the solutions.

When x>2 :

  (x+1) is positive, (x2) is positive and (x+3) is positive

Thus, (x+1)(x2)(x+3)>0

So, x>2 is one of the solutions.

Combining all the solutions, we have:

  x=3 or 3<x<1 or x=1 or x=2 or x>2

Thus, the solution set is 3x1 or x2

Drawing the above inequality on a real number line, we have:

  Single Variable Calculus: Concepts and Contexts, Enhanced Edition, Chapter A, Problem 22E , additional homework tip  2

Conclusion:

Hence, the solution set of the inequality is 3x1 or x2 and the solution set on a real number line -

  Single Variable Calculus: Concepts and Contexts, Enhanced Edition, Chapter A, Problem 22E , additional homework tip  3

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