Problem 52 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Let \(f(x)=2 x^{2}, g(x)=x+1\), and $h(x)=\frac{1}{x} . If what is written is an expression, simplify it. If it is an equation, solve it. $$ g(x) h(f(x))+f(x) h(g(x)) $$

Step-by-Step Solution

Answer

The simplified form of the given expression \(g(x) h(f(x))+f(x) h(g(x))\) is \(\frac{1}{2x} + \frac{1}{2x^2} + \frac{2x^2}{x+1}\).

Step 1: Substitute the Functions into the Expression

The first step is to substitute \(f(x)=2 x^{2}\), \(g(x)=x+1\), and \(h(x)=\frac{1}{x}\) into the given expression \(g(x) h(f(x))+f(x) h(g(x))\). The substitution gives us \((x + 1)\left(\frac{1}{2x^2}\right) + (2x^2)\left(\frac{1}{x+1}\right)\).

Step 2: Simplify the Expression

Now we simply carry out the operations in the expression. It simplifies to \(\frac{x+1}{2x^2} + \frac{2x^2}{x+1}\).

Step 3: Further Simplify the Expression

Next, split the first term into two and simplify to obtain \(\frac{1}{2x} + \frac{1}{2x^2} + \frac{2x^2}{x+1}\).

Step 4: Combine Like Terms

This expression cannot be further simplified, hence: \(g(x) h(f(x))+f(x) h(g(x)) = \frac{1}{2x} + \frac{1}{2x^2} + \frac{2x^2}{x+1}\)

Key Concepts

Algebraic Expression Simplification

When working with algebraic expressions, simplification involves rewriting an expression in a more concise form without changing its value. Simplification techniques include factorization, combining like terms, and canceling common factors. In the provided exercise, the aim is to simplify the complex expression formed by different function compositions. This involves a few straightforward steps:

• Substitute the given functions into the expression.
• Perform operations like multiplication, factoring, or division carefully.
• Look for opportunities to combine like terms or simplify fractions.

In this exercise, the original complex expression is broken down into simpler parts through strategic substitutions and calculations. Eventually, the expression is reduced to a mix of fraction terms, showing the clearest form possible without altering the mathematical equivalence.

Rational Functions

Rational functions are ratios of polynomial expressions. They have the form \( \frac{p(x)}{q(x)} \), where both \( p(x) \) and \( q(x) \) are polynomials, and \( q(x) eq 0 \). These functions can be simplified by reducing the fraction, but it is crucial to ensure that the denominator never equals zero, as this would make the function undefined.
In the step-by-step solution provided, each term of the expression is a rational function. By substituting and simplifying, we observe rational expressions like \( \frac{x+1}{2x^2} \) and \( \frac{2x^2}{x+1} \). In simplifying rational functions:

• Cancel out common factors when possible.
• Simplify the numerator and the denominator separately, then together.
• Check for undefined values, i.e., the values that make the denominator zero.

The exercise showcases how rational expressions add complexity to an equation but can be simplified systematically. Understanding rational functions is crucial in algebra as they often appear in various branches of mathematics.

Polynomial Functions

A polynomial function consists of variables raised to non-negative integer powers and combined using addition, subtraction, and multiplication. Polynomials form the building blocks of many mathematical concepts, as they are easy to manipulate, differentiate, and integrate.
In this exercise, the initial functions \( f(x) = 2x^2 \) and \( g(x) = x+1 \) are polynomials. Each plays a significant role in building the expression to simplify. Polynomials are particularly useful because:

• Their degree determines their shape and number of roots or solutions.
• They can be added, subtracted, and multiplied to form more complex expressions.
• They act as a foundation for more advanced functions like rational or trigonometric functions.

Recognizing and handling polynomial functions as part of function composition prepares students for more extensive applications and ensures a solid algebraic foundation. In situations like the one presented, understanding the nature of each polynomial involved can significantly aid in simplifying the expression efficiently.