/*! 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 19 Find the intervals on which \(f\... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 19

Find the intervals on which \(f\) is increasing and decreasing. Superimpose the graphs of \(f\) and \(f^{\prime}\) to verify your work. $$f(x)=(x-1)^{2}$$

Short Answer

Expert verified
Answer: The function \(f(x) = (x-1)^2\) is decreasing on the interval \((-\infty, 1)\) and increasing on the interval \((1, \infty)\).

Step by step solution

01

Find the derivative of the function

Calculate the derivative of \(f(x) = (x-1)^2\) with respect to \(x\). Use the power rule: $$f'(x) = 2(x-1)$$
02

Determine the intervals where \(f'(x)\) is positive and negative

To find these intervals, we need to find the value of \(x\) for which \(f'(x)=0\): $$2(x-1)=0$$ Solving for \(x\), we get: $$x=1$$ Now, consider two cases: Case 1: \(x < 1\) Substitute \(x\) with a number smaller than 1 (for example, 0) into \(f'(x)\) to check its sign: $$f'(0) = 2(0-1) = -2$$ Since \(f'(0)\) is negative, \(f'(x) < 0\) for any \(x\) in the interval \((-\infty, 1)\). Case 2: \(x > 1\) Substitute \(x\) with a number greater than 1 (for example, 2) into \(f'(x)\) to check its sign: $$f'(2) = 2(2-1) = 2$$ Since \(f'(2)\) is positive, \(f'(x) > 0\) for any \(x\) in the interval \((1, \infty)\).
03

Determine where the function is increasing and decreasing

Using the signs of \(f'(x)\) we found in Step 2: - The function is decreasing when \(f'(x) < 0\), which occurs in the interval \((-\infty, 1)\). - The function is increasing when \(f'(x) > 0\), which occurs in the interval \((1, \infty)\). So \(f(x)\) is decreasing on the interval \((-\infty, 1)\) and increasing on the interval \((1, \infty)\).
04

Superimpose the graphs of \(f\) and \(f'\)

To verify the answer, plot the graphs of \(f(x) = (x-1)^2\) and \(f'(x) = 2(x-1)\) on the same coordinate plane. On the graph, you should be able to see: - The original function \(f(x)\) has a minimum at \(x=1\). - The function is decreasing to the left of \(x=1\) and increasing to the right of \(x=1\). - The derivative \(f'(x)\) changes its sign at \(x=1\) from negative to positive, which confirms the function's behavior. By analyzing the graphs, you can confirm that \(f(x)\) is decreasing on the interval \((-\infty, 1)\) and increasing on the interval \((1, \infty)\), which matches our findings in Steps 2 and 3.

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

Use analytical methods to evaluate the following limits. $$\lim _{x \rightarrow \infty}(\sqrt{x-2}-\sqrt{x-4})$$

Use analytical methods to evaluate the following limits. $$\lim _{t \rightarrow \pi / 2^{+}} \frac{\tan 3 t}{\sec 5 t}$$

Sketch the graph of a function that is continuous on \((-\infty, \infty)\) and satisfies the following sets of conditions. $$\begin{aligned}&f^{\prime \prime}(x)>0 \text { on }(-\infty,-2) ; f^{\prime \prime}(-2)=0 ; f^{\prime}(-1)=f^{\prime}(1)=0\\\&f^{\prime \prime}(2)=0 ; f^{\prime}(3)=0 ; f^{\prime \prime}(x)>0 \text { on }(4, \infty)\end{aligned}$$

Use analytical methods to evaluate the following limits. $$\lim _{n \rightarrow \infty} \frac{1+2+\cdots+n}{n^{2}}( \text {Hint}:$$ $$\left.1+2+\cdots+n=\frac{n(n+1)}{2}.\right)$$

A mass oscillates up and down on the end of a spring. Find its position \(s\) relative to the equilibrium position if its acceleration is \(a(t)=\sin (\pi t),\) and its initial velocity and position are \(v(0)=3\) and \(s(0)=0,\) respectively.
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Statistics

Read Explanation

Decision Maths

Read Explanation

Discrete Mathematics

Read Explanation

Applied Mathematics

Read Explanation

Mechanics Maths

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.