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Chapter 4: Problem 9

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

Short Answer

Expert verified
Global minimum: -1 at x = -1 and x = 1. Global maximum: 8 at x = 4.

Step by step solution

01

Address Absolute Value

First, address the absolute value in the function. The function given is \( f(x) = x^2 - 2|x| \). This can be broken down into two cases based on \( x \): one for \( x \geq 0 \) where \( |x| = x \), and the other for \( x < 0 \) where \( |x| = -x \). Thus, the function can be expressed as: - For \( x \geq 0 \), \( f(x) = x^2 - 2x \)- For \( x < 0 \), \( f(x) = x^2 + 2x \)
02

Find Critical Points

Differentiate the function formulas for different cases: 1. For \( x \geq 0 \): - \( f'(x) = 2x - 2 \) - Setting \( f'(x) = 0 \), we find the critical point at \( x = 1 \).2. For \( x < 0 \): - \( f'(x) = 2x + 2 \) - Setting \( f'(x) = 0 \), we find the critical point at \( x = -1 \).
03

Evaluate Function at Critical Points

Evaluate the function \( f(x) \) at the critical points and at the endpoints of the interval \([-3, 4]\):- \( f(-3) = (-3)^2 + 2(-3) = 9 - 6 = 3 \)- \( f(-1) = (-1)^2 + 2(-1) = 1 - 2 = -1 \)- \( f(1) = (1)^2 - 2(1) = 1 - 2 = -1 \)- \( f(4) = (4)^2 - 2(4) = 16 - 8 = 8 \)
04

Determine Global Extremes

Identify the highest and lowest values from the evaluations:- At \( x = -1 \) and \( x = 1 \), the value is \( -1 \).- At \( x = -3 \), the value is \( 3 \).- At \( x = 4 \), the value is \( 8 \).Thus, the global minimum is \( -1 \) and the global maximum is \( 8 \).

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Key Concepts

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

Global Maximum
In calculus, the global maximum is defined as the highest point over the entire domain of a function. For a function on a closed interval, this means finding the greatest output value, or "peak".
To locate a global maximum, we check critical points within the interval and the endpoints of the interval itself. It's essential to evaluate the function at these potential points to identify which one yields the highest value.
  • Critical points are where the derivative of the function is zero or undefined.
  • The endpoints of the interval are the boundary values over which we explore the function's behavior.
In our case, evaluating the function at critical points and endpoints revealed that the global maximum value of the function was attained at the endpoint of the interval with a value of 8.
Global Minimum
The global minimum of a function is the lowest point over its entire interval. Finding this point involves similar steps to finding the global maximum.
Within a closed interval, compute the value of the function at critical points and at the boundary endpoints of the interval. The smallest value from these results is the global minimum.
  • Evaluating at each critical point provides insight into local behavior.
  • End values offer a complete picture of the function's range.
For the given function, the calculation showed a global minimum value of -1. This was achieved at the critical points where the function dipped to its lowest.
Critical Points
Critical points are locations on a function's graph where the derivative equals zero or doesn't exist. These points indicate where the slope of the tangent to the curve is horizontal or undefined.
To identify them, we differentiate the function and solve for zero. In scenarios involving an absolute value function, consider its piecewise definition to address different behaviors in distinct intervals.
For our function, differentiating each piece separately, we found critical points at:
  • For \( x \geq 0 \), the critical point is at \( x = 1 \).
  • For \( x < 0 \), the critical point is at \( x = -1 \).
These critical points were essential in evaluating the global behavior of the function within its interval.
Absolute Value Function
An absolute value function is a piecewise function described by two different expressions. The formula differs based on whether the input is positive or negative.
This function returns only the non-negative value of its input. This straightforward feature requires careful consideration when conducting calculus operations, like differentiation.
In our exercise, we wrote the absolute value function as:
  • \( f(x) = x^2 - 2x \) when \( x \geq 0 \) because \( |x| = x \).
  • \( f(x) = x^2 + 2x \) when \( x < 0 \) because \( |x| = -x \).
This bifurcation allowed us to properly handle each segment of the function for further analysis, a crucial step to find critical points, and eventually, the global minimum and maximum.

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Most popular questions from this chapter

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?

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. If \(f^{\prime}(x)\) is continuous and \(f(x)\) has no critical points, then \(f\) is everywhere increasing or everywhere decreasing.

In a \(19^{\text {th }}\) century sea-battle, the number of ships on each side remaining \(t\) hours after the start are given by \(x(t)\) and \(y(t) .\) If the ships are equally equipped, the relation between them is \((x(t))^{2}-(y(t))^{2}=c,\) where \(c\) is a positive constant. The battle ends when one side has no ships remaining. (a) If, at the start of the battle, 50 ships on one side oppose 40 ships on the other, what is the value of \(c ?\) (b) If \(y(3)=16,\) what is \(x(3) ?\) What does this represent in terms of the battle? (c) There is a time \(T\) when \(y(T)=0 .\) What does this \(T\) represent in terms of the battle? (d) At the end of the battle, how many ships remain on the victorious side? (e) At any time during the battle, the rate per hour at which \(y\) loses ships is directly proportional to the number of \(x\) ships, with constant of proportionality k. Write an equation that represents this, Is \(k\) positive or negative? (f) Show that the rate per hour at which \(x\) loses ships is directly proportional to the number of \(y\) ships, with constant of proportionality \(k\) (g) Three hours after the start of the battle, \(x\) is losing ships at the rate of 32 ships per hour. What is \(k ?\) At what rate is \(y\) losing ships at this time?

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

The temperature, \(T,\) in \(^{\circ} \mathrm{C},\) of a yam put into a \(200^{\circ} \mathrm{C}\) oven is given as a function of time, \(t,\) in minutes, by $$T=a\left(1-e^{-k t}\right)+b$$ (a) If the yam starts at \(20^{\circ} \mathrm{C},\) find \(a\) and \(b\) (b) If the temperature of the yam is initially increasing at \(2^{\circ} \mathrm{C}\) per minute, find \(k\).
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