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Chapter 1: Problem 2

(a) find the simplified form of the difference quotient and then (b) complete the following table. $$ \begin{array}{|c|l|l|} \hline x & h & \frac{f(x+h)-f(x)}{h} \\ \hline 5 & 2 & \\ \hline 5 & 1 & \\ \hline 5 & 0.1 & \\ \hline 5 & 0.01 & \\ \hline \end{array} $$ $$ f(x)=4 x^{2} $$

Short Answer

Expert verified
The simplified form is \(8x + 4h\), and the table values are 48, 44, 40.4, and 40.04.

Step by step solution

01

Write the Expression for the Difference Quotient

The difference quotient of a function is given by \( \frac{f(x+h) - f(x)}{h} \). We need to substitute the given function \( f(x) = 4x^2 \) into this formula.
02

Substitute the Function into the Difference Quotient

Substitute \( f(x) = 4x^2 \) into the expression to get \( f(x+h) = 4(x+h)^2 \). So, the difference quotient becomes: \( \frac{4(x+h)^2 - 4x^2}{h} \).
03

Expand the Expression \(4(x+h)^2\)

Expand \( (x+h)^2 \) to get \( x^2 + 2xh + h^2 \). Thus, \( 4(x+h)^2 = 4(x^2 + 2xh + h^2) = 4x^2 + 8xh + 4h^2 \).
04

Simplify the Difference Quotient Expression

Substitute the expanded form back into the difference quotient: \( \frac{4x^2 + 8xh + 4h^2 - 4x^2}{h} \). Simplify this to \( \frac{8xh + 4h^2}{h} \).
05

Factor and Simplify Further

Factor out \( h \) from the numerator: \( \frac{h(8x + 4h)}{h} \). Cancel \( h \) from the numerator and the denominator to get \( 8x + 4h \).
06

Calculate the Values for Each \( h \) in the Table

Now use the simplified result \( 8x + 4h \) to complete the table.1. For \( x = 5, h = 2: 8(5) + 4(2) = 40 + 8 = 48 \).2. For \( x = 5, h = 1: 8(5) + 4(1) = 40 + 4 = 44 \).3. For \( x = 5, h = 0.1: 8(5) + 4(0.1) = 40 + 0.4 = 40.4 \).4. For \( x = 5, h = 0.01: 8(5) + 4(0.01) = 40 + 0.04 = 40.04 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Calculus
Calculus is a branch of mathematics that explores rates of change and the accumulation of quantities. It's divided primarily into two parts: differential calculus and integral calculus.
One fundamental concept in calculus is the derivative, which represents the rate of change of a function concerning a variable. This allows us to understand how a quantity increases or decreases over time or space.
Calculus uses various mathematical techniques to perform these analyses, such as limits and infinitesimals. The process of finding a derivative is called differentiation. Derivatives have practical applications in fields like physics, engineering, economics, and biology, where they are used to model real-world situations.
- Derivative finds the slope of a function at any given point. - Integrals, another part of calculus, measure the total accumulation of a quantity. Let's see how the difference quotient fits into this picture.
Derivative Approximation
Derivative approximation involves estimating the value of a derivative based on known values of a function.
The difference quotient is a primary tool used here. It is an expression that gives the average rate of change of a function over a small interval and is central to understanding derivatives in calculus.
The form of the difference quotient is:\[\frac{f(x+h) - f(x)}{h}\]This formula estimates the derivative of a function at a given point as \( h \) approaches zero. For the function given in the exercise, \( f(x) = 4x^2 \), the exercise uses this formula to approximate the derivative at specific intervals.- The smaller the \( h \), the closer the approximation to the actual derivative.- As seen in the exercise, as \( h \) decreases, the expression \( 8x + 4h \) provides a better approximation of the derivative.This process is essential to understand how functions behave locally.
Function Analysis
Function analysis involves studying the properties and behavior of mathematical functions.
This helps in understanding the nature of various mathematical models both theoretically and practically.
In analyzing functions, we often look for key characteristics like continuity, limits, and derivatives to better understand the behavior of a function over its domain.
  • Continuity ensures that a function doesn't have any unexpected breaks or jumps.
  • Limits help in understanding the behavior of functions at specific points or as inputs approach specific values.
  • Derivatives allow us to identify how a function's values change concerning changes in its inputs.
Examining the difference quotient, for instance, gives us an insight into the function's derivative, which tells us about the slope of the function at any point. In this way, function analysis allows us to deeply understand and predict the behavior of complex systems based on their mathematical representations.

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Most popular questions from this chapter

Find dy/dx. Each function can be differentiated using the rules developed in this section, but some algebra may be required beforehand. $$ y=(x+3)(x-2) $$

Find dy/dx. Each function can be differentiated using the rules developed in this section, but some algebra may be required beforehand. $$ y=\frac{x^{5}-x^{3}}{x^{2}} $$

The cost of sending a large envelope via U.S. first-class mail in 2014 was \(\$ 0.98\) for the first ounce and \(\$ 0.21\) for each additional ounce (or fraction thereof). (Source: www.usps.com.) If \(x\) represents the weight of a large envelope, in ounces, then \(p(x)\) is the cost of mailing it, where $$ \begin{array}{l} p(x)=\$ 0.98, \text { if } \quad 0 < x \leq 1, \\ p(x)=\$ 1.19, \text { if } \quad 1 < x \leq 2, \\ p(x)=\$ 1.40, \text { if } 2 < x \leq 3, \end{array} $$ and so on, up through 13 ounces. The graph of \(p\) is shown below. Using the graph of the postage function, find each of the following limits, if it exists. $$ \lim _{x \rightarrow 3.4} p(x) $$

Find dy/dx. Each function can be differentiated using the rules developed in this section, but some algebra may be required beforehand. $$ y=\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^{2} $$

The function given by \(R(x)=11.74 x^{1 / 4}\) can be used to approximate the maximum range \(\mathrm{R}(x)\) in miles, of an ARSR-3 surveillance radar with a peak power of \(x\) watts (W). (Source: Introduction to RADAR Techniques, Federal Aviation Administration, \(1988 .\). a) Find the rate at which the maximum radar range changes as peak power increases from \(40,000 \mathrm{~W}\) to \(60,000 \mathrm{~W}\) b) Find \(\frac{R(60,000)-R(50,000)}{60,000-50,000}\). What does this rate represent?
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