Question

Show that $$ |x| \leq 1 \Rightarrow\left|x^{4}+\frac{1}{2} x^{3}+\frac{1}{4} x^{2}+\frac{1}{8} x+\frac{1}{16}\right|<2 $$

Short answer

The expression is less than 2 for \(|x| \leq 1\).

Step-by-step solution

Understanding the function

We need to analyze the expression \( x^4 + \frac{1}{2}x^3 + \frac{1}{4}x^2 + \frac{1}{8}x + \frac{1}{16} \) and determine its maximum value. The domain given is \( |x| \leq 1 \). We will consider the contribution of each term when \( x \) is at its boundary values, which are \( x = 1 \) and \( x = -1 \).

Calculate the expression at boundary x = 1

Substitute \( x = 1 \) into the polynomial: \[ 1^4 + \frac{1}{2}(1)^3 + \frac{1}{4}(1)^2 + \frac{1}{8}(1) + \frac{1}{16} = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16}. \] Calculate the sum: \[ 1 + 0.5 + 0.25 + 0.125 + 0.0625 = 1.9375. \]

Calculate the expression at boundary x = -1

Substitute \( x = -1 \) into the polynomial: \[ (-1)^4 + \frac{1}{2}(-1)^3 + \frac{1}{4}(-1)^2 + \frac{1}{8}(-1) + \frac{1}{16} = 1 - \frac{1}{2} + \frac{1}{4} - \frac{1}{8} + \frac{1}{16}. \] Calculate the sum: \[ 1 - 0.5 + 0.25 - 0.125 + 0.0625 = 0.6875. \]

Consider the impact of x values between -1 and 1

Within the interval \( -1 \leq x \leq 1 \), the polynomial's contribution from each term remains bounded as seen in the previous calculations. Since the polynomial at the boundaries are below 2, by continuity and the nature of polynomial functions, it won't exceed 2 anywhere within the exact domain.

Conclusion

Having evaluated the polynomial at the critical boundary points \( x = 1 \) and \( x = -1 \), and considering the polynomial's continuous nature, we conclude that the expression for any \( x \) such that \( |x| \leq 1 \) satisfies: \[ \left|x^4 + \frac{1}{2} x^3 + \frac{1}{4} x^2 + \frac{1}{8} x + \frac{1}{16} \right| < 2. \]

Key concepts

Polynomial Boundaries

When dealing with polynomial boundaries, you essentially analyze the behavior of a polynomial at certain key points known as boundaries. In this exercise, the domain is defined by the inequality \(|x| \leq 1\). The boundaries of this domain are where \(x\) takes values of -1 and 1. These are the critical points where the polynomial can potentially have its extreme values, either a maximum or minimum.

By evaluating the polynomial at these boundary points, you can gain insight into its behavior over the entire domain. For the given polynomial, \(x^4 + \frac{1}{2}x^3 + \frac{1}{4}x^2 + \frac{1}{8}x + \frac{1}{16}\), we found the values at \(x = 1\) and \(x = -1\). The results were 1.9375 and 0.6875, respectively.

Evaluating these boundary points helps confirm the inequality \|x^4 + \frac{1}{2}x^3 + \frac{1}{4}x^2 + \frac{1}{8}x + \frac{1}{16}\| < 2 for the entire domain \(|x| \leq 1\). In this scenario, if the polynomial does not exceed the threshold at these points, it’s unlikely to do so within the domain itself.

Continuous Functions

Continuous functions play a crucial role in mathematical analysis. A continuous function is one that does not have any breaks, jumps, or undefined points in its domain.

For this polynomial exercise, the fact that the polynomial is a continuous function is key to proving the inequality. Polynomials are inherently continuous over their domain of definition—an interval like \([-1, 1]\) in this case.

• A continuous function ensures that if it is below a certain value at the endpoints of an interval, it will also remain below that value between those endpoints.

Thus, for the polynomial in this problem, observing the function's behavior at \(x = 1\) and \(x = -1\) indicates its maximum possible behavior across the entire interval. Since its maximum at the boundaries is less than 2, continuity implies it must be less than 2 throughout.

Polynomial Evaluation

Evaluating a polynomial involves substituting specific values into the equation and simplifying to find the resulting value. This concept is essential when determining the behavior of a polynomial at given points, especially boundary ones.

For instance, in this exercise, substituting \(x = 1\) and \(x = -1\) allows us to find:

• At \(x = 1\): The result of the expression \(1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16}\) simplifies to 1.9375.
• At \(x = -1\): The expression simplifies to \(1 - \frac{1}{2} + \frac{1}{4} - \frac{1}{8} + \frac{1}{16}\), which totals 0.6875.

These evaluations confirm the polynomial remains well below 2 at these critical boundary points, hinting at its behavior over the whole continuous interval \([-1, 1]\). This insight underpins evaluating whether inequalities like \(|x^4 + \frac{1}{2} x^3 + \frac{1}{4} x^2 + \frac{1}{8} x + \frac{1}{16}| < 2\) are satisfied across a defined domain.

Inequality Solutions

Solving inequalities involving polynomials requires analyzing both algebraic behavior and graphical interpretation. It relates closely to understanding boundary behavior and continuity, as seen in this exercise.

The original problem asks to show that \(|x^4 + \frac{1}{2} x^3 + \frac{1}{4} x^2 + \frac{1}{8} x + \frac{1}{16}| < 2\) for \(|x| \leq 1\).

• First, solving involves substituting boundary values into the polynomial, as these showcase potential extremes on a given interval.
• The solution amalgamates these evaluations, with continuous function properties to infer behavior through the entire domain.

Unlike equations, inequalities offer a range of solutions due to providing bounds rather than exact values. Using boundary evaluations here confirmed the polynomial value never reaches or exceeds 2, achieving the desired conclusion for all \(x\) in \([-1, 1]\). Understanding and solving such inequalities equip students with critical skills for exploring broader topics in calculus and analysis.