/*! 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 27 Find all critical numbers by han... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find all critical numbers by hand. If available, use graphing technology to determine whether the critical number represents a local maximum, local minimum or neither. $$f(x)=\left|x^{2}-1\right|$$

Short Answer

Expert verified
The function \(f(x)=|x^{2}-1|\) has a critical number at \(x = 0\). Upon evaluation, this critical number is neither a local maximum nor a local minimum, but a point of inflection.

Step by step solution

01

Define the function

Since we're given a function that involves absolute values, we will define it in pieces depending on the value of \(x\). So: \(f(x) = x^2 - 1\) for \(x ≥ 1\) and \(f(x) = 1 - x^2\) for \(x ≤ 1\).
02

Find the derivative

The next step is to find the derivative of \(f(x)\) for both conditions. For \(x ≥ 1\), \(f'(x) = 2x\). For \(x ≤ 1\), \(f'(x) = -2x\).
03

Find where the derivative is zero or does not exist

This step is crucial for finding the critical numbers. For \(f'(x) = 2x\), the derivative is 0 when \(x = 0\). Similarly, for \(f'(x) = -2x\), the derivative is also 0 when \(x = 0\). Hence, the critical number where the derivative is zero is \(x = 0\). As this is a polynomial function, its derivative exists for all real numbers, so there are no critical numbers where the derivative does not exist.
04

Determine if the critical number is a local maximum, local minimum, or neither

To find this out, consider the function values just to the right and left of \(x = 0\). Just to the left (\(x = -0.1\)), the function \(f(x) = 1.01\). Slightly to the right (\(x = 0.1\)), it is also \(f(x) = 1.01\). Hence, \(x = 0\) is neither a local maximum nor a local minimum. It's a point of inflection.

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.

Absolute Value Functions
Absolute value functions involve expressions within vertical bars, indicating that the value taken is always non-negative. They’re represented as |x|, where x could be any real number. This implies that the graph reflects the input x as itself when x is positive or zero, and as -x when x is negative, causing a characteristic 'V' shape in their graphs.

For the given function, when we deal with the absolute value of a function like \(f(x)=|x^2-1|\), it's crucial to consider both the inside function \(x^2-1\) flipping its sign depending on whether the inside expression is positive or negative. This is why the function is defined in pieces for \(x ≥ 1\) and \(x ≤ 1\). Understanding these pieces allows us to analyze the graph and differentiate effectively.
Derivatives
Derivatives represent the rate at which a function is changing at any given point and are foundational in calculus for identifying critical numbers, which highlight important features of the graph like slopes of tangents and rates of change. Derivatives are typically denoted as \(f'(x)\) for a function \(f(x)\).

In the given exercise, computing the derivative of the absolute value function creates two separate scenarios depending on the sign of the inside expression. For \(x ≥ 1\), the function's derivative is straightforward, \(f'(x) = 2x\), and for \(x ≤ 1\), the derivative is negative, \(f'(x) = -2x\), owing to the 'flip' caused by the absolute value. Knowing how to calculate and interpret derivatives is essential to understanding the behavior of a function.
Local Maximum and Minimum
Local maximum and minimum points are high and low spots in a function where the function value peaks or troughs relative to nearby points. These points are particularly interesting because they represent extremal values in a certain interval.

When analyzing \(f(x) = x^2 - 1\) for \(x ≥ 1\) and \(f(x) = 1 - x^2\) for \(x ≤ 1\), we use derivatives to find where the slope of the tangent line is horizontal—the function's rate of change being zero. This is generally done by setting the derivative equal to zero. In this context, the critical number \(x = 0\) is checked for being a local max or min by comparing the function values around it. As seen, \(x = 0\) has the same value for slight perturbations to the left and right, indicating it is neither, which can be a transitional point, known as an inflection point.
Point of Inflection
A point of inflection is a point on the curve of the graph of a function where the curve changes from being concave (curved upwards) to convex (curved downwards), or vice versa. In simpler terms, it signifies a change in the direction of curvature.

For the function \(f(x) = |x^2 - 1|\), the critical number found is \(x = 0\). Despite this point not being a local max or min, it's relevant because the concavity of the function changes at this point. The rate of change does not 'speed up' or 'slow down' in the usual sense of a maximum or minimum; instead, it 'switches direction'. This is a more sophisticated concept involving the second derivative, but even with first derivative analysis, students can observe this shift in the graph’s curvature. Understanding inflection points deepens one's comprehension of a function's geometry.

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

The table shows the coefficient of friction \(\mu\) of ice as a function of temperature. The lower \(\mu\) is, the more "slippery" the ice is. Estimate \(\mu^{\prime}(C)\) at \((a) C=-10\) and \((b) C=-6 .\) If skating warms the ice, does it get easier or harder to skate? Briefly explain. $$\begin{array}{|c|c|c|c|c|c|c|} \hline^{\circ} C & -12 & -10 & -8 & -6 & -4 & -2 \\ \hline \mu & 0.0048 & 0.0045 & 0.0043 & 0.0045 & 0.0048 & 0.0055 \\ \hline \end{array}$$

Sand is poured into a conical pile with the height of the pile equalling the diameter of the pile. If the sand is poured at a constant rate of \(5 \mathrm{m}^{3} / \mathrm{s}\), at what rate is the height of the pile increasing when the height is 2 meters?

Suppose a \(6-f t\) -tall person is 12 ft away from an 18 -fi-tall lamppost (see the figure). If the person is moving away from the lamppost at a rate of \(2 \mathrm{ft} / \mathrm{s}\), at what rate is the length of the shadow changing? (Hint: Show that \(\left.\frac{x+s}{18}=\frac{s}{6} \text { . }\right)\)

In the titration of a weak acid and strong base, the pH is given by \(c+\ln \frac{x}{1-x},\) where \(f\) is the fraction \((0

Suppose that a species reproduces as follows: with probability \(p_{0},\) an organism has no offspring; with probability \(p_{1},\) an organism has one offspring; with probability \(p_{2},\) an organism has two offspring and so on. The probability that the species goes extinct is given by the smallest nonnegative solution of the equation \(p_{0}+p_{1} x+p_{2} x^{2}+\cdots=x\) (see Sigmund's Games of Life). Find the positive solutions of the equations \(0.1+0.2 x+0.3 x^{2}+0.4 x^{3}=x\) and \(0.4+0.3 x+0.2 x^{2}+0.1 x^{3}=x .\) Explain in terms of species going extinct why the first equation has a smaller solution than the second.

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.