Problem 8 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Let \(f(x)=x(x+1), g(x)=x^{3}+2 x^{2}+x\). (a) Simplify the following. i. \(f(x)+g(x)\) ii. \(\frac{f(x)}{g(x)}\) iii. \(\frac{g(x)}{f(x)}\) iv. \(\frac{[f(x)]^{2}}{g(x)}\) (b) Solve \(x f(x)=g(x)\).

Step-by-Step Solution

Answer

The simplified expressions for \(f(x) + g(x)\), \(\frac{f(x)}{g(x)}\), \(\frac{g(x)}{f(x)}\), \(\frac{[f(x)]^2}{g(x)}\) are \(x^3 + 3x^2 + 2x\), \(\frac{1}{x}\), \(x\) and \(x^2 + 2x + 1\) respectively. The solutions to the equation \(x f(x)=g(x)\) are x = 0 and x = -1.

Step 1: Simplify \(f(x) + g(x)\)

Begin by recalling that adding functions simply translates to adding respective terms. \(f(x) + g(x) = x(x+1) + (x^3 + 2x^2 + x) = x^3 + 3x^2 + 2x\)

Step 2: Simplify \(\frac{f(x)}{g(x)}\) and \(\frac{g(x)}{f(x)}\)

The division operation in this context means that the function in the numerator is divided by the function in the denominator. i. \(\frac{f(x)}{g(x)} = \frac{x(x+1)}{x^3 + 2x^2 + x} = \frac{x(x+1)}{x(x^2 + 2x + 1)} = \frac{1}{x}\) as the \(x(x+1)\) are canceled out. ii. \(\frac{g(x)}{f(x)} = \frac{x^3 + 2x^2 + x}{x(x+1)} = \frac{x^3 + 2x^2 + x}{x(x^2 + x)} = x\) as we can divide the numerator and the denominator by \(x(x+1)\)

Step 3: Simplify \(\frac{[f(x)]^2}{g(x)}\)

\(\frac{[f(x)]^2}{g(x)} = \frac{(x^2 + 2x + 1)^2}{x^3 + 2x^2 + x} = \frac{(x^2 + 2x + 1)^2}{x(x^2 + 2x + 1)} = (x^2 + 2x + 1)\) as we can divide the numerator and the denominator by \(x(x^2 + 2x + 1)\)

Step 4: Solve \(x f(x)=g(x)\)

We'll equate \(x f(x)\) to \(g(x)\) and proceed to find the roots of the equation. \(x (x(x+1)) = x^3 + 2x^2 + x = 0\). This reduces to a quadratic equation and can be solved as follows: \(x^2 + x = 2x^2 + 2x\). Rearranging terms, we get x^2 + x = 0 so x(x + 1) = 0. Therefore, the solutions are x = 0 and x = -1.

Key Concepts

Function Operations

Understanding function operations allows us to manipulate functions in various ways. In the context of calculus problems such as the exercise provided, operations include addition and division of functions.

For instance, to add functions like \(f(x)\) and \(g(x)\), we simply add the corresponding terms. This is what we see when \(f(x)\) is added to \(g(x)\) resulting in \(x^3 + 3x^2 + 2x\). It's a process of combining like terms, very much akin to standard polynomial addition.

Division of functions, however, is a bit more intricate. It involves rewriting the numerator and denominator so that common terms can be canceled out. In the given exercise, when dividing \(f(x)\) by \(g(x)\), we notice that the term \(x(x+1)\) appears in both the numerator and the denominator, allowing us to simplify the expression to \(\frac{1}{x}\).

Simplifying Expressions

Simplifying mathematical expressions is a cornerstone of algebra and calculus. The goal here is to reduce complex equations to their simplest forms for ease of understanding and solving. It entails combining like terms, factoring, expanding expressions, and canceling out common factors.

In our textbook example, the expression \(\frac{[f(x)]^2}{g(x)}\) was simplified to \((x^2 + 2x + 1)\) by canceling the common factor, which in this case is \(x(x^2 + 2x + 1)\). This is a great demonstration of simplifying expressions; it's all about recognizing patterns and reducing computation whenever possible.

Polynomial Division

Polynomial division can often seem daunting, but it is just an extension of basic division we use in arithmetic, applied to polynomials. When dividing one polynomial by another, we are looking for how many times the divisor can 'fit' into the dividend, or in other words, what multiple of the divisor gives us the dividend.

The step-by-step solution offered for the exercise includes an example of polynomial division where \(g(x)\) is divided by \(f(x)\). Once both functions are expressed in factored form, it's clearer to see the common terms that can be divided out, leaving us with \(x\) as the simplified expression for \(\frac{g(x)}{f(x)}\). Understanding this process is critical for solving complex calculus problems that involve polynomial functions.

Solving Equations

Solving equations is the process of finding the values, called roots, that make the equation true. For linear equations, this involves isolating the variable on one side of the equation. However, in the context of polynomial equations, which often appear in calculus problems, this might involve factoring and using the Zero Product Property.

In the exercise, to solve the equation \(x f(x) = g(x)\), we first need to express both sides as polynomials and then equate them to zero. Once in this form, we can factor the equation and find the values of \(x\) that satisfy the conditions. For the given problem, the roots came out to be \(x = 0\) and \(x = -1\), which are where the function intersects the x-axis on a graph.