/*! 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 17 Identify the coordinates of any ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Identify the coordinates of any local and absolute extreme points and inflection points. Graph the function. $$y-x^{4}-2 x^{2}-x^{2}\left(x^{2}-2\right)$$

Short Answer

Expert verified
The function is the constant line \(y = 0\) with no extrema or inflection points.

Step by step solution

01

Simplify the Given Function

The given function is \[ y = x^4 - 2x^2 - x^2(x^2 - 2) \]First, let's distribute and simplify the terms:\[ y = x^4 - 2x^2 - x^4 + 2x^2 \]Combine like terms:\[ y = 0 \]Thus, the function simplifies to a constant, \( y = 0 \).
02

Analyze Critical Points and Extrema

Since the function is a constant \( y = 0 \), there are no variable-dependent critical points, local extrema, or absolute extrema as the value of \( y \) does not change for any \( x \). The entire graph is a horizontal line, so every point is both locally and absolutely constant.
03

Determine Inflection Points

An inflection point requires a change in the concavity of the graph. However, since \( y = 0 \) is a horizontal line without any curvature, it does not have inflection points either.
04

Graph the Function

The function \( y = 0 \) is represented graphically as a horizontal line on the cartesian plane. It runs along the x-axis, intersecting it at every point of the x-axis.

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 91Ó°ÊÓ!

Key Concepts

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

Critical Points
In calculus, we often want to understand where a function might change direction on its graph. This is where critical points become important.
Critical points occur at values of the independent variable (typically x) where the derivative of a function equals zero or is undefined. These are places where the graph has a horizontal tangent or a sharp turn.
For a given function, you find the derivative and solve for x where this derivative equals zero. This process helps identify potential spots of change like peaks (maximums) or valleys (minimums). In the case of our exercise with the function simplifying to a constant \[ y = 0 \], there are no variable-dependent critical points because the derivative, being zero across all x, never changes. Essentially, there's no place where the slope of the function becomes horizontal because it's constantly flat. Keep in mind these key points when identifying critical locations:
  • Find the derivative.
  • Solve for where this derivative equals zero.
  • Consider all possible undefined points in the derivative as potential critical points.
Understanding these basics helps to determine changes in a graph's direction.
Local Extrema
Local extrema refer to the highest or lowest points in a particular region or interval of a graph. These can be either local maxima or minima, and they typically occur at critical points we previously discussed.
To find a local extremum, examine the sign changes of the derivative on either side of a critical point:
  • Local Maximum: The function changes from increasing to decreasing.
  • Local Minimum: The function changes from decreasing to increasing.
In our simplified function
\( y = 0 \), there are no highs or lows since it's a constant line on the y-axis. There's no change in value, meaning no local extrema exist.
The concept of local extrema is vital for understanding the behavior of more complex functions where such peaks and troughs can offer significant insights into the function's overall shape and behavior. However, for a constant function like ours, every point is essentially the same without any peaks or dips.
Inflection Points
Inflection points are where the graph of a function changes its curvature. This means the function shifts from being concave up (shaped like a cup) to concave down (shaped like a cap), or vice versa. It's where the second derivative equals zero.
To determine these points for any function:
  • Calculate the second derivative.
  • Find values for x where this second derivative changes sign.
An inflection point indicates a change in the rate of increase or decrease, highlighting a fundamental shift in the graph's direction.
For our constant function,
\( y = 0 \), there's no change in curvature. The graph remains a flat line. Thus, it has no inflection points as the second derivative is zero throughout.
In more variable cases, locating inflection points is crucial for mapping true behavior, revealing where concavity swaps and helping in accurate graph sketching.

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

Solve the initial value problems. $$\begin{aligned} &y^{(4)}=-\sin t+\cos t\\\ &y^{\prime \prime \prime}(0)=7, \quad y^{\prime \prime}(0)=y^{\prime}(0)=-1, \quad y(0)=0 \end{aligned}$$

You are to construct an open rectangular box with a square base and a volume of \(6 \mathrm{m}^{3}\). If material for the bottom costs \(\$ 60 / \mathrm{m}^{2}\) and material for the sides costs \(\$ 40 / \mathrm{m}^{2}\), what dimensions will result in the least expensive box? What is the minimum cost?

Find the curve \(y=f(x)\) in the \(x y\) -plane that passes through the point (9,4) and whose slope at each point is \(3 \sqrt{x}\).

The hammer and the feather When Apollo 15 astronaut David Scott dropped a hammer and a feather on the moon to demonstrate that in a vacuum all bodies fall with the same (constant) acceleration, he dropped them from about \(1.2 \mathrm{m}\) above the ground. The television footage of the event shows the hammer and the feather falling more slowly than on Earth, where, in a vacuum, they would have taken only half a second to fall the \(1.2 \mathrm{m}\). How long did it take the hammer and feather to fall \(1.2 \mathrm{m}\) on the moon? To find out, solve the following initial value problem for \(s\) as a function of \(t .\) Then find the value of \(t\) that makes \(s\) equal to \(0 .\). Differential equation: \(\frac{d^{2} s}{d t^{2}}=-1.6 \mathrm{m} / \mathrm{s}^{2}\) Initial conditions: \(\frac{d s}{d t}=0\) and \(s=1.2\) when \(t=0\)

Find the most general antiderivative or indefinite integral. You may need to try a solution and then adjust your guess. Check your answers by differentiation. $$\int\left(\sin 2 x-\csc ^{2} x\right) d x$$

See all solutions

Recommended explanations on Math Textbooks

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.