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Chapter 4: Problem 5

Find the global maximum and minimum for the function on the closed interval. $$f(x)=x^{4}-8 x^{2}, \quad-3 \leq x \leq 1$$

Short Answer

Expert verified
Global maximum is 9 at \(x = -3\); global minimum is -16 at \(x = -2\).

Step by step solution

01

Identify Critical Points

To find critical points, first compute the derivative of the function. The derivative of \(f(x) = x^4 - 8x^2\) is \(f'(x) = 4x^3 - 16x\). Set the derivative equal to zero to find the critical points: \[4x^3 - 16x = 0\] Factor the expression: \[4x(x^2 - 4) = 0\] This results in \(x = 0\), \(x = 2\), and \(x = -2\). Only \(x = 0\) and \(x = -2\) are within the interval \([-3, 1]\).
02

Evaluate at Endpoints and Critical Points

Calculate \(f(x)\) for \(x = -3\), \(x = 1\), and the critical points within the interval, \(x = 0\) and \(x = -2\). Calculate: - \(f(-3) = (-3)^4 - 8(-3)^2 = 81 - 72 = 9\) - \(f(1) = (1)^4 - 8(1)^2 = 1 - 8 = -7\) - \(f(0) = (0)^4 - 8(0)^2 = 0\) - \(f(-2) = (-2)^4 - 8(-2)^2 = 16 - 32 = -16\)
03

Determine Global Maximum and Minimum

Now, compare the values obtained: - \(f(-3) = 9\) - \(f(0) = 0\) - \(f(-2) = -16\) - \(f(1) = -7\) The highest value is 9 at \(x = -3\), so the global maximum is \(9\). The lowest value is -16 at \(x = -2\), so the global minimum is \(-16\).

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Key Concepts

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

Global Maximum
Finding the global maximum of a function involves identifying the point on a graph where the function reaches its highest value. In this exercise, after calculating the function values at both critical points and the endpoints of the closed interval, we determine that the global maximum is found at \(x = -3\) where \(f(x) = 9\).

These steps include evaluating:
  • The derivative to find critical points.
  • Function values at these critical points.
  • Function values at the interval's endpoints.
The global maximum value represents the highest peak on a graph within the given interval. Always ensure to check these points to identify where the maximum value occurs on a closed interval. Remember, only points within the interval are valid for consideration when determining maxima.
Global Minimum
Similar to finding a global maximum, identifying the global minimum involves locating the point where the function's value is the lowest within a given interval. In our specific problem, the function reaches its global minimum at \(x = -2\), where \(f(x) = -16\).

To locate a global minimum:
  • Evaluate the function at all critical points within the interval.
  • Check the function's values at the endpoints of the interval.
The global minimum represents the deepest valley on the function's graph within the specified interval. This process involves analyzing all potential points where the function can achieve a minimum value, ensuring that all critical and endpoint values are considered.
Derivative
The derivative of a function is a tool that helps us understand the behavior of the function, particularly its slopes or rates of change. In this exercise, we computed the derivative of \(f(x) = x^4 - 8x^2\), resulting in \(f'(x) = 4x^3 - 16x\).

The derivative is crucial because:
  • Setting it to zero helps find critical points.
  • Critical points indicate where maximum and minimum values may occur.
In solving for critical points, we set the derivative equal to zero, factor it, and solve the resulting equation. This reveals points of interest including both local extremes and possible global extrema within a specified interval.
Mastery of derivatives is essential for studying calculus and real-world applications, including physics and engineering.
Closed Interval
A closed interval in calculus is a range of values on which a function is defined and includes both its endpoints. For this exercise, the closed interval is \([-3, 1]\). It signifies that the evaluation of the function should include the boundary points \(-3\) and \(1\).

Why closed intervals matter:
  • Endpoints may contain global maxima or minima.
  • They limit the domain of interest for our solutions.
  • Ensure inclusivity of boundary points when evaluating function behavior.
In practical terms, a closed interval ensures a complete consideration of the function's behavior across specified limits. It is an essential concept when solving optimization problems and guarantees that we find true extremes by considering the whole defined range.

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Most popular questions from this chapter

Are the statements in Problems true or false for a function \(f\) whose domain is all real numbers? If a statement is true,explain how you know. If a statement is false, give a counterexample. A local minimum of \(f\) occurs at a critical point of \(f\).

A gas station stands at the intersection of a north-south road and an east- west road. A police car is traveling toward the gas station from the east, chasing a stolen truck which is traveling north away from the gas station. The speed of the police car is 100 mph at the moment it is 3 miles from the gas station. At the same time, the truck is 4 miles from the gas station going 80 mph. At this moment: (a) Is the distance between the car and truck increasing or decreasing? How fast? (Distance is measured along a straight line joining the car and the truck.) (b) How does your answer change if the truck is going 70 mph instead of 80 mph?

Let \(f(x)=a x+b / x .\) Suppose \(a\) and \(b\) are positive. What happens to \(f(x)\) as \(b\) increases? (a) The critical points move further apart. (b) The critical points move closer together. (c) The critical values move further apart. (d) The critical values move closer together.

Investigate the given two parameter family of functions. Assume that \(a\) and \(b\) are positive. (a) Graph \(f(x)\) using \(b=1\) and three different values for \(a\). (b) Graph \(f(x)\) using \(a=1\) and three different values for \(b\). (c) In the graphs in parts (a) and (b), how do the critical points of \(f\) appear to move as \(a\) increases? As \(b\) increases? (d) Find a formula for the \(x\) -coordinates of the critical point(s) of \(f\) in terms of \(a\) and \(b\). $$f(x)=\sqrt{b-(x-a)^{2}}$$

Find formulas for the functions described. A function of the form \(y=a \sin \left(b t^{2}\right)\) whose first critical point for positive \(t\) occurs at \(t=1\) and whose derivative is 3 when \(t=2\).
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