# To solve the below inequality in terms of intervals and illustrate the solution set on the real number line - x 3 + 3 x &lt; 4 x 2

### Single Variable Calculus: Concepts...

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

### Single Variable Calculus: Concepts...

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

#### Solutions

Chapter A, Problem 24E
To determine

## To solve the below inequality in terms of intervals and illustrate the solution set on the real number line -   x3+3x<4x2

Expert Solution

The solution of the inequality is x<0 or 1<x<3 and the solution set on a real number line -

### Explanation of Solution

Given: Inequality: x3+3x<4x2

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 x3+3x<4x2

Simplifying the above inequality, we have:

x3+3x<4x2

x34x2+3x<0x(x24x+3)<0x(x1)(x3)<0

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

When x<0 :

x is negative, (x1) is negativeand (x3) is negative.

Thus, x(x1)(x3)<0

So, x<0 isone of the solutions.

When 0<x<1 :

x is positive, (x1) is negativeand (x3) is negative.

Thus, x(x1)(x3)>0

So, 0<x<1 is not one of the solutions.

When 1<x<3 :

x is positive, (x1) is positive and (x3) is negative.

Thus, x(x+1)(x1)<0

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

When x>3 :

x is positive, (x1) is positive and (x3) is positive.

Thus, x(x1)(x3)>0

So, x>3 is not one of the solutions.

Combining all the solutions, we have the solution set as:

x<0 or 1<x<3

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

Conclusion:

Hence, the solution of the inequality is x<0 or 1<x<3 and the solution set on a real number line -

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