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Chapter 6: Problem 23

Use a graphing utility to estimate the absolute maximum and minimum values of \(f\), if any, on the stated interval, and then use calculus methods to find the exact values. $$f(x)=\left(x^{2}-1\right)^{2}:(-\infty,+\infty)$$

Short Answer

Expert verified
Absolute minimum: \(f = 0\) at \(x = 1\) and \(x = -1\). No absolute maximum.

Step by step solution

01

Analyzing the Function

The given function is \( f(x) = (x^2 - 1)^2 \). This function is defined for all real numbers, hence the interval of interest is \(( -\infty, +\infty)\). Understanding this function's behavior will help in identifying turning points where potential maximum and minimum values could exist.
02

Graphing the Function

Using a graphing utility, plot the function \( f(x) = (x^2 - 1)^2 \). The graph appears symmetric around the y-axis, showing potential local maxima and minima. The graph decreases to infinity in both directions (x -> +\infty and x -> -\infty), probably reaching some minimum value somewhere between.
03

Finding Critical Points Using Calculus

To find the critical points, first compute the derivative of the function: \( f'(x) = \frac{d}{dx} [(x^2 - 1)^2] = 2(x^2 - 1)(2x) = 4x(x^2 - 1) \). Set \( f'(x) = 0 \) to find the critical points: \( 4x(x^2 - 1) = 0 \). This gives solutions \( x = 0, x = 1, x = -1 \).
04

Evaluating Function at Critical Points

Calculate \( f(x) \) at each critical point: - \( f(0) = (0^2 - 1)^2 = 1 \). - \( f(1) = (1^2 - 1)^2 = 0 \). - \( f(-1) = ((-1)^2 - 1)^2 = 0 \). Among these, \( x = 1 \) and \( x = -1 \) yield a minimum value of 0.
05

Determining Absolute Extrema

The minimum values calculated from the critical points (0 at \( x = 1 \) and \( x = -1 \)) are the absolute minimum values since the function approaches infinity for \( x \to +\infty \) or \( x \to -\infty \). Therefore, the function does not have an absolute maximum value.

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Key Concepts

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

Graphing Utility
A graphing utility is a digital tool that helps visualize functions by plotting their graphs. Utilizing a graphing utility can simplify the process of understanding the behavior of complex functions like \( f(x) = (x^2 - 1)^2 \). By inputting this function into the utility, you can quickly see its symmetry and how it behaves over a large interval.
This specific function is defined over all real numbers and displays symmetry around the y-axis.
  • The graph drops towards infinity as \( x \) approaches both positive and negative infinity.
  • This visual tool identifies approximate locations of local maxima and minima, acting as a guide for further calculus evaluations.
Using such utilities not only saves time but also aids in honing your analytical skills as you set the stage for calculus-based explorations.
Absolute Maximum and Minimum
The terms "absolute maximum" and "absolute minimum" refer to the highest and lowest values a function can achieve over its entire domain or on a specified interval. In the exercise case, it focuses on finding these values of \( f(x) = (x^2 - 1)^2 \) on \(( -\infty, +\infty)\).
As visualized in the graph and confirmed by calculus:
  • The absolute minimum value is 0, which occurs at \( x = 1 \) and \( x = -1 \).
  • There's no absolute maximum as the function continues to rise without bound as \( x \) moves towards both infinities.

Identifying these extremes helps understand the full range of the function's output and provides key insights, especially when examining real-world applications of mathematical models.
Critical Points
Critical points are points on the graph of a function where its derivative is zero or undefined. These points are essential in determining where local maxima, minima, or saddle points might occur. For the function \( f(x) = (x^2 - 1)^2 \), you start by finding its critical points:
  • Calculate the derivative: \( f'(x) = 4x(x^2 - 1) \).
  • Set the derivative to zero: \( 4x(x^2 - 1) = 0 \).
  • Solve for \( x \): you find \( x = 0, x = 1, x = -1 \).

These solutions indicate potential extrema points:
  • \( x = 1 \) and \( x = -1 \) are where the function's lowest value occurs, as evaluated by later calculations.
  • \( x = 0 \) provides another possible point for examination.

These critical points give significant hints for evaluating the function more thoroughly when seeking absolute extremas.
Derivative
The derivative of a function is a fundamental concept in calculus, representing its rate of change. Finding the derivative is key in locating critical points and analyzing the nature of the function at those points. For \( f(x) = (x^2 - 1)^2 \), you calculate:
  • The derivative is \( f'(x) = 4x(x^2 - 1) \), derived using the chain rule and basic differentiation principles.
  • This derivative indicates where the velocity or slope of the function's graph is zero, suggesting potential critical points.

In essence, the derivative is an invaluable tool that allows us to dissect the behavior of functions methodically. Delving into derivatives helps ensure a complete understanding of how functions behave鈥攁n essential aspect of mastering calculus.

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Use a graphing utility to estimate the absolute maximum and minimum values of \(f\), if any, on the stated interval, and then use calculus methods to find the exact values. $$f(x)=\sin (\cos x) ;[0,2 \pi |$$
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