Chapter 1: Problem 68
Show that $$ |x| \leq 2 \Rightarrow\left|\frac{x^{2}+2 x+7}{x^{2}+1}\right| \leq 15 $$
Short Answer
Expert verified
The expression does not exceed 15 for all \(|x| \leq 2\).
Step by step solution
01
Understand the Problem Statement
We need to prove that when the absolute value of \(x\) is less than or equal to 2, the expression \(\left|\frac{x^{2}+2x+7}{x^{2}+1}\right|\) is less than or equal to 15.
02
Investigate the Expression
The expression in question is \(\frac{x^{2} + 2x + 7}{x^{2} + 1}\). We will evaluate this expression for values of \(x\) where \(|x| \leq 2\), i.e., the interval \([-2, 2]\).
03
Determine the Range of the Numerator
In the numerator, \(x^2 + 2x + 7\), for \(|x| \leq 2\), the largest possible value happens at the endpoints. Calculating at \(x = 2\), we have \(2^2 + 2(2) + 7 = 4 + 4 + 7 = 15\). Similarly for \(x = -2\), the value is the same because \((x^2 + 2x + 7)\) is symmetric around zero.
04
Evaluate the Denominator
In the denominator \(x^2 + 1\), for \(x = 2\) and \(x = -2\), both result in \(2^2 + 1 = 5\). The smallest value for \(x = 0\) results in \(0^2 + 1 = 1\), meaning the denominator does not become zero within the interval.
05
Calculate the Expression at Critical Points
Evaluate the expression at the critical points \(x = 2, 0, -2\). At each critical point, the value of the expression \(\frac{x^{2}+2x+7}{x^{2}+1}\) is either positive or negative but remains at or below 15.
06
Synthesizing the Information
The maximum value of the expression is at the edges, \(x = 2\) or \(x = -2\), giving \(\left|\frac{15}{5}\right| = 3\). Since 3 < 15, the condition holds for all points in the interval \([-2, 2]\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Absolute Value
Absolute value refers to the distance a number is from zero on the number line, without considering direction. For any real number \(x\), the absolute value is denoted as \(|x|\). This definition implies that \(|x|\) is always non-negative. For instance, \(|3| = 3\) and \(|-3| = 3\).
In the context of inequalities such as \(|x| \leq 2\), this means that \(x\) can be any number between -2 and 2, inclusive. We can express this range as the interval \([-2, 2]\). In solving problems involving absolute values, it's essential to consider these intervals since any value of \(x\) within this interval will maintain the inequality.
Absolute values are crucial in determining bounds within problem-solving, especially when linked with functions or expressions. They act as a lens to keep solutions within specified limits, ensuring the function behaves in a predictable manner.
In the context of inequalities such as \(|x| \leq 2\), this means that \(x\) can be any number between -2 and 2, inclusive. We can express this range as the interval \([-2, 2]\). In solving problems involving absolute values, it's essential to consider these intervals since any value of \(x\) within this interval will maintain the inequality.
Absolute values are crucial in determining bounds within problem-solving, especially when linked with functions or expressions. They act as a lens to keep solutions within specified limits, ensuring the function behaves in a predictable manner.
Rational Expressions
A rational expression is a fraction where both the numerator and the denominator are polynomials. For example, the expression \(\frac{x^2 + 2x + 7}{x^2 + 1}\) is a rational expression. It's a ratio of the polynomial \(x^2 + 2x + 7\) in the numerator to another polynomial \(x^2 + 1\) in the denominator.
Handling rational expressions involves understanding their behavior, particularly where the denominator could potentially be zero, which would make the expression undefined. However, in our problem, \(x^2 + 1\) in the denominator is always positive within the given interval \([-2, 2]\). This guarantees the expression is valid throughout.
When dealing with inequalities involving rational expressions, it's important to evaluate the expression at critical points - points where the function's behavior could change, such as at the ends of the interval. In our example, we checked \(x = -2, 0, \) and \(2\).
This evaluation ensured the expression remains valid and below the specified bound, which is critical for proving inequalities.
Handling rational expressions involves understanding their behavior, particularly where the denominator could potentially be zero, which would make the expression undefined. However, in our problem, \(x^2 + 1\) in the denominator is always positive within the given interval \([-2, 2]\). This guarantees the expression is valid throughout.
When dealing with inequalities involving rational expressions, it's important to evaluate the expression at critical points - points where the function's behavior could change, such as at the ends of the interval. In our example, we checked \(x = -2, 0, \) and \(2\).
This evaluation ensured the expression remains valid and below the specified bound, which is critical for proving inequalities.
Calculus Problem Solving
Calculus often involves finding maxima and minima of functions, a technique that is essential in proving inequalities like the one in our problem. In this exercise, we determined the maximum value of the rational expression \(\frac{x^2 + 2x + 7}{x^2 + 1}\) within the interval \([-2, 2]\).
To solve this, evaluating the function at the critical points within this interval was necessary. These include where the derivative of the expression could be zero (indicating a possible maxima or minima), and specifically at the boundaries \(x = -2\) and \(x = 2\).
In this context, calculus helps identify these critical points by providing tools like derivatives to see where changes occur. But more straightforward problem-solving can often involve direct computation, such as checking the expression values at these critical points directly. This avoids more complex calculus computations when unnecessary, offering a more elementary yet effective method of solving.
Problem-solving in calculus often requires this critical analysis to predict the behavior over an interval correctly. By determining these values, you ensure the function satisfies the inequality across the specified range.
To solve this, evaluating the function at the critical points within this interval was necessary. These include where the derivative of the expression could be zero (indicating a possible maxima or minima), and specifically at the boundaries \(x = -2\) and \(x = 2\).
In this context, calculus helps identify these critical points by providing tools like derivatives to see where changes occur. But more straightforward problem-solving can often involve direct computation, such as checking the expression values at these critical points directly. This avoids more complex calculus computations when unnecessary, offering a more elementary yet effective method of solving.
Problem-solving in calculus often requires this critical analysis to predict the behavior over an interval correctly. By determining these values, you ensure the function satisfies the inequality across the specified range.
