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Chapter 1: Problem 16

(a) find the simplified form of the difference quotient and then (b) complete the following table. $$ \begin{array}{|c|l|l|} \hline x & h & \frac{f(x+h)-f(x)}{h} \\ \hline 5 & 2 & \\ \hline 5 & 1 & \\ \hline 5 & 0.1 & \\ \hline 5 & 0.01 & \\ \hline \end{array} $$ $$ f(x)=x^{2}+4 x-3 $$

Short Answer

Expert verified
The simplified difference quotient is \(2x + h + 4\). Table values: 16, 15, 14.1, 14.01.

Step by step solution

01

Understand the Function

The given function is \(f(x) = x^2 + 4x - 3\). You are tasked with finding the difference quotient \(\frac{f(x+h) - f(x)}{h}\) for the function, and simplifying it.
02

Find \(f(x+h)\)

Start by substituting \(x + h\) into the function \(f(x)\). This means replacing every instance of \(x\) with \((x+h)\). The expression becomes \((x+h)^2 + 4(x+h) - 3\).
03

Expand \(f(x+h)\)

Expand \((x+h)^2\) to get \(x^2 + 2xh + h^2\), and \(4(x+h)\) to get \(4x + 4h\). Combine all expanded terms to find \[f(x+h) = x^2 + 2xh + h^2 + 4x + 4h - 3\].
04

Calculate the Difference Quotient

Subtract \(f(x)\) from \(f(x+h)\): \[ \begin{align*} & \left( x^2 + 2xh + h^2 + 4x + 4h - 3 \right) - \left( x^2 + 4x - 3 \right)\ &= 2xh + h^2 + 4h. \end{align*} \] Simplify by canceling like terms, then divide by \(h\): \(\frac{2xh + h^2 + 4h}{h} = 2x + h + 4\).
05

Simplify the Difference Quotient

The simplified form of the difference quotient is \(2x + h + 4\). This is the algebraic expression you will substitute values into for different \(h\).
06

Substitute into the Table

Now, substitute \(x = 5\) into the simplified form \(2x + h + 4\) for each \(h\): - For \(h = 2\), \(2(5) + 2 + 4 = 16\).- For \(h = 1\), \(2(5) + 1 + 4 = 15\).- For \(h = 0.1\), \(2(5) + 0.1 + 4 = 14.1\).- For \(h = 0.01\), \(2(5) + 0.01 + 4 = 14.01\).
07

Complete the Table

Fill in the table with these calculated values:\[\begin{array}{|c|c|c|}\hlinex & h & \frac{f(x+h) - f(x)}{h} \\hline5 & 2 & 16 \5 & 1 & 15 \5 & 0.1 & 14.1 \5 & 0.01 & 14.01 \\hline\end{array}\]

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Key Concepts

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

Simplifying Algebraic Expressions
Algebraic expressions can appear daunting at first glance. However, simplifying them is a crucial skill in mathematics, especially when dealing with functions and difference quotients. Simplification lets us transform complex expressions into more manageable forms. Let鈥檚 break down the process.
  • **Identify Similar Terms:** Group terms that have the same variable raised to the same power. Like terms can be combined, reducing the number of terms in an expression.
  • **Perform Arithmetic Operations:** Apply basic operations like addition, subtraction, multiplication, or division to these like terms.
  • **Apply Distributive Property:** Use this property to remove parentheses and combine terms by expanding expressions, such as \(a(b+c) = ab + ac\).

In the exercise, simplifying the difference quotient \(\frac{f(x+h) - f(x)}{h}\) is key. After expanding \(f(x+h)\), identifying like terms like \(x^2\) and \(x,\) and then carefully simplifying it \(\text{to}\ 2x + h + 4\), is essential. Breaking it down simplifies our task and makes functions easier to work with.
Polynomial Functions
Polynomial functions are mathematical expressions involving a sum of powers in one or more variables with coefficients. Understanding them helps in various algebraic manipulations, including calculating the difference quotient.
  • **Form and Degree:** A polynomial function looks like \(a_nx^n + a_{n-1}x^{n-1} + \, ... \, + a_1x + a_0\), where \(a\)'s are coefficients and \(n\) is the degree of the polynomial. The degree determines the function's complexity regarding its number of roots and turns.
  • **Examples:** Linear polynomials, like \(3x + 1\), have a degree of 1, while quadratic polynomials, such as \(x^2 + 4x - 3\), have a degree of 2.

In this exercise, the function \(f(x) = x^2 + 4x - 3\) is quadratic. Recognizing its form enabled us to substitute and expand \(f(x+h)\) correctly. Knowing the structure of polynomial functions aids in further calculus study, especially in understanding changes in their behavior.
Derivatives
Derivatives serve as fundamental tools in calculus, revealing rates of change and trends of functions. They provide insights into how a function behaves at any given point and are closely linked with the difference quotient.
  • **Difference Quotient as a Foundation:** The expression \(\frac{f(x+h) - f(x)}{h}\) approximates the derivative, providing a way to estimate how much a function changes, near \(x\). As \(h\) approaches zero, it gives us a precise rate of change 鈥� the derivative at \(x\).
  • **Calculating Derivatives:** For simple polynomial functions, the process involves using power rules where the derivative of \(x^n\) is \(nx^{n-1}\).

By simplifying the difference quotient in this task, \(2x + h + 4\), and observing results for smaller and smaller \(h\), one estimates the derivative of the given polynomial function at \(x = 5\). This shows how the theoretical underpinnings of derivatives apply practically, enhancing our understanding of function meanings and applications.

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Most popular questions from this chapter

In \(t\) seconds, an object dropped from a certain height will fall \(s(t)\) feet, where $$ s(t)=16 t^{2} $$ a) Find \(s(5)-s(3)\). b) What is the average rate of change of distance with respect to time during the period from 3 to 5 sec? This is known as average velocity, or speed.

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 an equation of the tangent line to the graph of \(f(x)=x^{2}- sqrt{x}\). a) at (1,0) ; b) at (4,14) c) at (9,78) .

The equation $$ S(r)=\frac{1}{r^{4}} $$ can be used to determine the resistance to blood flow, \(S\). of a blood vessel that has radius \(r\), in millimeters (mm). a) Find the rate of change of resistance with respect to \(r\), the radius of the blood vessel. b) Find the resistance at \(r=1.2 \mathrm{~mm}\). c) Find the rate of change of \(S\) with respect to \(r\) when \(r=0.8 \mathrm{~mm}\)

On the moon, all free-fall distance functions are of the form \(s(t)=0.81 t^{2},\) where \(t\) is in seconds and \(s(t)\) is in meters. An object is dropped from a height of 200 meters above the moon. After \(t=2 \mathrm{sec}\) a) How far has the object fallen? b) How fast is it traveling? c) What is its acceleration? d) Explain the meaning of the second derivative of this free-fall function.
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