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Chapter 2: Problem 103

Use your calculator's absolute-value feature to graph the follow. ing functions and determine relative extrema and intervals over which the function is increasing or decreasing. State the \(x\) -values at which the derivative does not exist. $$f(x)=\left|x^{2}-1\right|$$

Short Answer

Expert verified
Relative minima at \(x = -1\) and \(x = 1\); increases on \( (-\frac{1}{ \sqrt{2}} , - 1)\), \( (0, 1) \); decreases on \((-1, 0)\), \( (1, \frac{1}{ \sqrt{2}} ) \); derivative does not exist at \(x = -1, 0, 1 \).

Step by step solution

01

- Graph the function

Use a graphing calculator to input the function \(f(x)=|x^2-1|\). Observe that the graph will be a 'W' shape, with minimum points at \(x = -1\) and \(x = 1\), and a vertex at \(x = 0\).
02

- Identify relative extrema

Relative minima occur at the points where the function changes from decreasing to increasing. For \(f(x)=|x^2-1|\), the relative minima are at \(x = -1\) and \(x = 1\), with \(f(-1) = 0\) and \(f(1) = 0\).
03

- Determine intervals of increase and decrease

The function decreases on the intervals \(-\frac{1}{ \sqrt{2}} , - 1)\) and \((0, 1)\) and increases on the intervals \((-1, 0)\) and \((1, \frac{1}{ \sqrt{2}} )\). This can be observed from the graph, where the slope of the line segment changes from negative to positive.
04

- Identify x-values at which the derivative does not exist

The derivative does not exist at points where the function has a cusp or a corner. For the function \(f(x)=|x^2-1|\), the derivative does not exist at \(x = -1\), \(x = 0\), and \(x = 1\). These points are where the absolute value function changes direction.

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Key Concepts

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

graphing functions
Graphing absolute value functions can be tricky but very visual and intuitive. For the function \(f(x) = |x^2 - 1|\), start by understanding the underlying quadratic function \(g(x) = x^2 - 1\). Graph the quadratic function as a parabola.

The absolute value function, \(f(x) = |x^2 - 1|\), will mirror parts of this parabola that fall below the x-axis to above the x-axis. The graph looks like a 'W' shape:
  • Vertex at \(x = 0\)
  • Minimum points at \(x = -1\) and \(x = 1\)
Use a graphing calculator for a clearer picture. Input the function and observe how it transforms.
relative extrema
Relative extrema refer to the highest or lowest points in a particular interval of a function. For \(f(x) = |x^2 - 1|\), identify the key points where the function has dips or peaks.

In this case, the function has relative minima at \( x = -1 \) and \( x = 1 \). At these points, the function changes its direction from decreasing to increasing.

Specifically, \( f(-1) = 0 \) and \( f(1) = 0 \). These are the lowest points in small surrounding intervals and are called relative minima.
intervals of increase and decrease
To determine where a function is increasing or decreasing, look at the slopes of its graph:
  • A positive slope means the function is increasing.
  • A negative slope means the function is decreasing.
For \(f(x) = |x^2 - 1|\):

  • The function decreases in the intervals \( (-\frac{1}{\text{ \sqrt{2}}}, -1) \) and \( (0, 1) \).
  • It increases in the intervals \( (-1, 0) \) and \( (1, \frac{1}{\text{ \sqrt{2}}}) \).
These intervals are identified by noting where the slopes change from negative to positive in the graph.
non-existent derivatives
Derivatives do not exist at points where a function has sharp corners or cusps. For \(f(x) = |x^2 - 1|\), the derivative does not exist at the following points:

  • \(x = -1\)
  • \(x = 0\)
  • \(x = 1\)
At these x-values, the graph has sudden changes in direction, creating vertices or corners. That's why the function lacks smoothness, making its derivative non-existent at these points.

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Most popular questions from this chapter

Find the rates of change of total revenue, cost, and profit with respect to time. Assume that \(R(x)\) and \(C(x)\) are in dollars. \(R(x)=50 x-0.5 x^{2}\) \(C(x)=10 x+3\) when \(x=10\) and \(d x / d t=5\) units per day

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Find the maximum profit and the number of units that must be produced and sold in order to yield the maximum profit. Assume that revenue, \(R(x),\) and \(\cos t, C(x)\) are in dollars. $$R(x)=50 x-0.5 x^{2}, \quad C(x)=10 x+3$$

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