Problem 80 - Calculus with Concepts in Calculus

Question

a. Show that \(x1\). b. Show that \(x^{2}

Step-by-Step Solution

Answer

a. \(x1\). b. \(x^2

Step 1: Understanding the inequality x<x^2

Consider the inequality \(x < x^2\). This means \(x - x^2 < 0\). Rewrite this as \(x(1-x) < 0\). To understand when this inequality holds, examine the factors \(x\) and \((1-x)\).

Step 2: Analyzing the factors separately for x 1

For \(x < 0\), \(x\) is negative and \((1-x)\) is positive since \(1 - x > 1\). Thus, \(x(1-x)<0\) is true. For \(x > 1\), \(x\) is positive and \((1-x)\) is negative since \(1-x < 0\). Thus \(x(1-x)<0\) is true as well.

Step 3: Conclusion for part (a)

For \(x < 0\) or \(x > 1\), \(x(1-x) < 0\). Therefore, \(x < x^2\) for \(x < 0\) or \(x > 1\).

Step 4: Understanding the inequality x^2<x

Consider the inequality \(x^2 < x\). This means \(x^2 - x < 0\). Factor this as \(x(x-1) < 0\). To find when this inequality holds, analyze the factors \(x\) and \((x-1)\).

Step 5: Analyzing the factors separately for 0<x<1

For \(0 < x < 1\), \(x\) is positive and \((x-1)\) is negative since \(x - 1 < 0\). Therefore, \(x(x-1)<0\) is true.

Step 6: Conclusion for part (b)

For \(0 < x < 1\), \(x(x-1) < 0\). Therefore, \(x^2 < x\) for \(0 < x < 1\).

Key Concepts

Quadratic Inequality

When dealing with quadratic inequalities, we are often examining mathematical statements that involve quadratic terms, such as the exercises given in the problem. For example, the inequality \(x < x^2\) is quadratic because of the term \(x^2\). The process involves finding intervals of \(x\) that satisfy the inequality.
This is done by transforming the inequality into a product of linear factors. For instance, by rewriting \(x < x^2\) as \(x - x^2 < 0\), and further simplifying this to \(x(1-x) < 0\), we can better understand and solve given inequalities.
Remember, quadratic inequalities can have up to two solutions, since quadratic equations can have up to two roots. Finding where the inequality holds involves checking intervals determined by these roots.

Factorization

Factorization is critical in solving quadratic inequalities. It involves breaking down an expression into a product of simpler expressions. For example, in the inequality \(x^2 < x\), rearranging gives \(x^2 - x < 0\), which can be factored as \(x(x-1) < 0\).
This step is essential because it reveals important information about the sign of the inequality.
Whenever you factor a quadratic expression, you expose the zero points or roots which are crucial to interval testing. These roots are where the expression could change signs, helping us determine on which intervals the inequality holds.

Algebraic Inequality Analysis

Once you've factorized a quadratic inequality, like \(x(1-x) < 0\) or \(x(x-1) < 0\), the next step is to analyze these factors. Inequality analysis involves determining where a product of factors is less than, greater than, or equal to zero.
To do this, consider the sign of each factor over different intervals. For instance, for \(x(x-1)<0\):

• When \(x < 0\), \(x\) is negative and \(x-1\) is also negative, resulting in their product being positive.
• When \(0 < x < 1\), \(x\) is positive and \(x-1\) is negative, resulting in a negative product.
• When \(x > 1\), both factors are positive, resulting in a positive product.

The inequality holds where the product is negative. Thus, thorough analysis based on interval testing is key in solving algebraic inequalities.

Previous

Problem 79

Next

Problem 81

Other exercises in this chapter

Problem 78

If \(x^{3}>125\), is it necessarily true that \(x>5\) ? Explain.

View solution

Problem 79

Is \(1 / x

View solution

Problem 81

Use the definition of absolute value to prove the following. a. \(|a b|=|a||b|\) b. \(-|b| \leq b \leq|b|\) c. \(|a-b|=|b-a|\)

View solution

Problem 83

Show that \(|a+b|=|a|+|b|\) if and only if \(a b \geq 0\) (which means that \(a=0, b=0\), or \(a\) and \(b\) have the same sign).

View solution

Key Concepts

Quadratic Inequality

When dealing with quadratic inequalities, we are often examining mathematical statements that involve quadratic terms, such as the exercises given in the problem. For example, the inequality \(x < x^2\) is quadratic because of the term \(x^2\). The process involves finding intervals of \(x\) that satisfy the inequality.
This is done by transforming the inequality into a product of linear factors. For instance, by rewriting \(x < x^2\) as \(x - x^2 < 0\), and further simplifying this to \(x(1-x) < 0\), we can better understand and solve given inequalities.
Remember, quadratic inequalities can have up to two solutions, since quadratic equations can have up to two roots. Finding where the inequality holds involves checking intervals determined by these roots.

Factorization

Factorization is critical in solving quadratic inequalities. It involves breaking down an expression into a product of simpler expressions. For example, in the inequality \(x^2 < x\), rearranging gives \(x^2 - x < 0\), which can be factored as \(x(x-1) < 0\).
This step is essential because it reveals important information about the sign of the inequality.
Whenever you factor a quadratic expression, you expose the zero points or roots which are crucial to interval testing. These roots are where the expression could change signs, helping us determine on which intervals the inequality holds.

Algebraic Inequality Analysis

Once you've factorized a quadratic inequality, like \(x(1-x) < 0\) or \(x(x-1) < 0\), the next step is to analyze these factors. Inequality analysis involves determining where a product of factors is less than, greater than, or equal to zero.
To do this, consider the sign of each factor over different intervals. For instance, for \(x(x-1)<0\):

• When \(x < 0\), \(x\) is negative and \(x-1\) is also negative, resulting in their product being positive.
• When \(0 < x < 1\), \(x\) is positive and \(x-1\) is negative, resulting in a negative product.
• When \(x > 1\), both factors are positive, resulting in a positive product.

The inequality holds where the product is negative. Thus, thorough analysis based on interval testing is key in solving algebraic inequalities.