Problem 52 - Precalculus: Functions and Graphs

Question

Use the graph of \(f\) to approximate the remaineder if \(f\) is divided by \(x-0.21\). $$f(x)=3.33 x^{6}-2.5 x^{5}+6.9 x^{3}-4.1 x^{2}+1.22 x-6.78$$

Step-by-Step Solution

Answer

The remainder is approximately \(-6.56\).

Step 1: Understanding the Remainder Theorem

The Remainder Theorem states that if a polynomial \( f(x) \) is divided by \( x - c \), the remainder of this division is \( f(c) \). This means that to find the remainder of \( f(x) \) divided by \( x-0.21 \), we need to evaluate \( f(0.21) \).

Step 2: Substitute into the Polynomial

To find \( f(0.21) \), substitute \( 0.21 \) into the polynomial: \[ f(0.21) = 3.33(0.21)^6 - 2.5(0.21)^5 + 6.9(0.21)^3 - 4.1(0.21)^2 + 1.22(0.21) - 6.78 \]

Step 3: Calculate Exponents and Products

First, calculate each term involving exponents and products: \( 3.33 \times (0.21)^6 \), \( -2.5 \times (0.21)^5 \), \( 6.9 \times (0.21)^3 \), \( -4.1 \times (0.21)^2 \), \( 1.22 \times 0.21 \).

Step 4: Simplify and Sum the Terms

Evaluate the results from the previous calculations and sum them up: \( f(0.21) = 3.33(0.21)^6 + (-2.5)(0.21)^5 + 6.9(0.21)^3 + (-4.1)(0.21)^2 + 1.22(0.21) - 6.78 \).

Step 5: Approximate the Remainder

After calculations, approximate the value of \( f(0.21) \). This value is the remainder of the division of \( f(x) \) by \( x-0.21 \).

Key Concepts

Polynomial Division

Polynomial division is a fundamental concept when dealing with polynomials, similar to dividing numbers. It's used to divide one polynomial by another, usually resulting in a quotient and a remainder.
To perform polynomial division, you need to know both the dividend, which is the polynomial you're dividing, and the divisor, the polynomial by which you divide.

Consider polynomial division analogous to long division of numbers, where:

• The dividend is broken down into parts, starting from the highest degree term.
• Each part is divided by the leading term of the divisor.
• Products and differences are handled step-by-step until all parts of the dividend are accounted for.
• The remainder is what's left after the last division step.

In the context of the Remainder Theorem, the essence of polynomial division is simplified: instead of going through complex division steps, you only need to evaluate the polynomial at a specific point to find the remainder. This makes finding remainders both intuitive and precise.

Polynomial Evaluation

Polynomial evaluation is a straightforward technique to determine the output of a polynomial function for a given input. It involves substituting a particular number into the variables of a polynomial equation and simplifying to find the result.
In the context of the Remainder Theorem, evaluating the polynomial at a given value is crucial because this evaluation directly provides the remainder when the polynomial is divided by a linear expression of the form \(x - c\).

For example, to evaluate the polynomial \( f(x) = 3.33x^6 - 2.5x^5 + 6.9x^3 - 4.1x^2 + 1.22x - 6.78 \) at \( x = 0.21 \):

• Substitute \( 0.21 \) into each occurrence of \( x \) in the polynomial.
• Calculate the power and product for each term.
• Add all the results together to obtain \( f(0.21) \).

This result represents both the value of the polynomial at \( x = 0.21 \) and the remainder of the division of \( f(x) \) by \( x - 0.21 \). It's a practical and efficient method for solving polynomial division problems when the divisor is linear.

Precalculus

Precalculus serves as a bridge between algebra and calculus, encompassing various mathematical concepts including polynomials, their properties, and functions. It aims to prepare students for studying calculus by building foundational skills in mathematics.

• Polynomials: Understanding polynomials, from basic operations to more advanced topics like the Remainder Theorem, is a vital part of precalculus.
• Remainder Theorem: This theorem simplifies the process of division by linking it with evaluation, which is a key topic in precalculus.
• Functions and Graphs: Analyzing graphs of polynomial functions helps students intuitively grasp the effects of different factors, building a visual understanding that supports calculus study.

Overall, precalculus integrates these concepts to ensure students grasp the nature of functions, their graphical representations, and operational management, setting a strong foundation for calculus and further mathematical courses.