/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 136 determine whether each statement... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 136

determine whether each statement makes sense or does not make sense, and explain your reasoning. I noticed that the difference quotient is always zero if \(f(x)=c,\) where \(c\) is any constant.

Short Answer

Expert verified
The statement makes sense. For a constant function \(f(x) = c\), the difference quotient is indeed always zero. This happens because the function value does not change with different x-values, thus making the slope of the secant line (difference quotient) equals zero.

Step by step solution

01

Understanding Difference Quotient

Before addressing the question, first, we need to understand what difference quotient is. The difference quotient for a function \(f(x)\) is given by the formula,\[\frac{f(x + h) - f(x)}{h}\]This compares the change in the function values, \(f(x + h) - f(x)\), with the change in x-values, \(h\). This is essentially the average rate of change or the slope of the secant line to the graph of \(f\) over the interval \(x\) to \(x + h\).
02

Testing the Claim for Constant Function

Now, let's test the claim in the problem for a constant function \(f(x)=c\). We are going to substitute \(f(x) = c\) into our formula and see what we get. \[\frac{f(x + h) - f(x)}{h} = \frac{c - c}{h} = \frac{0}{h} = 0\]No matter what the values of \(x\) and \(h\), the numerator will always be \(0\) because \(f(x + h) = f(x) = c\) for a constant function. Hence, the difference quotient will be \(0\), as dividing \(0\) by any non-zero number, \(h\), in this case, gives \(0\).
03

Explanation of the Results

The reasoning behind this result lies in the fact that a constant function don't change, hence the reason it's called constant. This means that, the outputs of the function are the same for any input value of \(x\). There's no rate of change as there's no change in the function value. This makes the difference quotient equals zero.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Complete the square and write the equation in standard form. Then give the center and radius of each circle and graph the equation. $$x^{2}+y^{2}-4 x-12 y-9=0$$

Determine whether each statement makes sense or does not make sense, and explain your reasoning. In graph of \((x-2)^{2}+(y+1)^{2}=16\) is my graph of \(x^{2}+y^{2}=16\) translated two units right and one unit down.

Find a. \((f \circ g)(x)\) b. \((g \circ f)(x)\) c. \((f \circ g)(2)\) d. \((g \circ f)(2)\) $$f(x)=x^{2}+1, g(x)=x^{2}-3$$

Express the given function \(h\) as \(a\) composition of two functions \(f\) and \(g\) so that \(h(x)=(f \circ g)(x)\). $$h(x)=\frac{1}{4 x+5}$$

graph both equations in the same rectangular coordinate system and find all points of intersection. Then show that these ordered pairs satisfy the equations. $$\begin{aligned} (x-2)^{2}+(y+3)^{2} &=4 \\ y &=x-3 \end{aligned}$$
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Probability and Statistics

Read Explanation

Decision Maths

Read Explanation

Discrete Mathematics

Read Explanation

Logic and Functions

Read Explanation

Statistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.