Question

Simplify the difference quotients\(\frac{f(x+h)-f(x)}{h}\) and \(\frac{f(x)-f(a)}{x-a}\) for the following functions. $$f(x)=x^{2}$$

Short answer

Question: Simplify the following difference quotients for the function \(f(x) = x^2\): 1. \(\frac{f(x+h)-f(x)}{h}\) 2. \(\frac{f(x)-f(a)}{x-a}\) Answer: 1. The simplified form of the first difference quotient is \(2x + h\). 2. The simplified form of the second difference quotient is \(x + a\).

Step-by-step solution

Substitute the function f(x) into the first difference quotient

First, we will substitute the function \(f(x) = x^2\) into the first difference quotient \(\frac{f(x+h)-f(x)}{h}\): $$\frac{f(x+h)-f(x)}{h} = \frac{(x+h)^2 - x^2}{h}$$

Expand and simplify the first difference quotient

Now, let's expand the numerator and simplify the expression as much as possible: \begin{align*} \frac{(x+h)^2 - x^2}{h} &= \frac{x^2 + 2xh + h^2 - x^2}{h}\\ &= \frac{2xh + h^2}{h} \end{align*}

Factor h from the first difference quotient

To further simplify the expression, let's factor out the \(h\) from the numerator: $$\frac{2xh + h^2}{h} = h\frac{2x + h}{h}$$

Simplify the first difference quotient completely

Finally, we can cancel the \(h\) from the numerator and the denominator: $$h\frac{2x + h}{h} = 2x + h$$ Thus, the simplified form of the first difference quotient is \(2x + h\). Now, let's move on to the second difference quotient.

Substitute the function f(x) into the second difference quotient

First, we will substitute the function \(f(x) = x^2\) into the second difference quotient \(\frac{f(x)-f(a)}{x-a}\): $$\frac{f(x)-f(a)}{x-a} = \frac{x^2 - a^2}{x-a}$$

Apply the difference of squares formula

Now, we can notice that the numerator is a difference of squares, so let's apply the difference of squares formula \(a^2 - b^2 = (a-b)(a+b)\) to simplify our expression: $$\frac{x^2 - a^2}{x-a} = \frac{(x-a)(x+a)}{x-a}$$

Simplify the second difference quotient completely

Lastly, we can cancel the \(x-a\) from the numerator and the denominator: $$\frac{(x-a)(x+a)}{x-a} = x+a$$ Thus, the simplified form of the second difference quotient is \(x + a\).

Key concepts

Difference Quotient

The difference quotient is an essential concept in calculus as it provides a method for finding the derivative of a function. It is used to measure how a function changes as its input changes. Specifically, it helps us understand the average rate of change of the function over an interval, which is crucial in determining instantaneous rates of change. Let's look at the two primary forms of the difference quotient:

• First Difference Quotient: \( \frac{f(x+h)-f(x)}{h} \)
• Second Difference Quotient: \( \frac{f(x)-f(a)}{x-a} \)

Each difference quotient plays a vital role in understanding the behavior of functions, and ultimately, leads us to the concept of derivatives when the limit of the difference quotient is taken as the interval approaches zero. The examples provided offer a practical exercise in applying these constructs to polynomial functions.

Simplification

Simplification is a process used to make mathematical expressions easier to work with. It often involves combining like terms or factoring and canceling terms. This is a useful skill when working with difference quotients because simplified expressions reveal insights more clearly. When simplifying, you will usually:

• Expand expressions: Opening up brackets and removing them, as seen with \((x+h)^2\).
• Combine like terms: Merging terms with common variables or coefficients.
• Factor expressions: Taking out common factors to streamline the calculation.
• Cancel out terms: Removing terms from both the numerator and the denominator, when applicable and valid.

In our example, after substituting the polynomial into the difference quotient, the process of simplification includes expanding \((x+h)^2\), canceling common terms, and reducing the expression to its simplest form. This is crucial for making the expression more manageable and in preparing for taking limits.

Polynomials

Polynomials are algebraic expressions that consist of variables raised to whole number powers, typically represented in the form \( a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0 \). They are fundamental in calculus because they are continuous and differentiable everywhere across their domain. This makes them ideal candidates for exploration through difference quotients.For example, the function we are working with, \(f(x) = x^2\), is a simple polynomial which is defined for all real numbers. Its quadratic nature simplifies the process when first substituting it into the difference quotient and expanding the terms. Polynomials provide a great foundation for understanding more complex calculus concepts, as their rates of change can be easily calculated and interpreted when using derivatives.

Limits

The concept of limits underpins much of calculus. A limit helps us understand the behavior of a function as it approaches a particular point, which is essential in defining derivatives and integrals.In the context of difference quotients, using limits involves examining what happens as \( h \to 0 \) or as \( x \to a \). The idea is to make the interval of change infinitesimally small, thereby turning the average rate of change into an instantaneous rate of change at a specific point. This is expressed as:

• \( \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \) for calculating the derivative of a function at a point.
• \( \lim_{x \to a} \frac{f(x)-f(a)}{x-a} \) to find the derivative at \( x = a \).

Understanding limits supports the accurate calculation of derivatives, enabling students to apply calculus to real-world problems, involving motion, growth, and change over time. In studying limits, we approach the foundational ideas of calculus that unlock the methods for analyzing and comprehending changes in variables.