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Chapter 4: Problem 16

Find the absolute maximum and minimum values of \(f\) on the given closed interval, and state where those values occur. $$ f(x)=|6-4 x| ;[-3,3] $$

Short Answer

Expert verified
Absolute maximum is 18 at \(x = -3\) and absolute minimum is 0 at \(x = 1.5\).

Step by step solution

01

Understand the function and interval

We need to find the absolute maximum and minimum of the function \(f(x) = |6 - 4x|\) on the closed interval \([-3, 3]\). Since \(|...|\) is an absolute value function, it's defined piecewise.
02

Determine the critical points

The critical points occur where the derivative exists and is zero or where the derivative does not exist. The function \(f(x) = |6 - 4x|\) can be rewritten as two linear functions: \(f(x) = 6 - 4x\) when \(4x \leq 6\) (i.e., \(x \leq 1.5\)), and \(f(x) = -6 + 4x\) when \(4x > 6\) (i.e., \(x > 1.5\)). The point \(x = 1.5\) is where the expression inside the absolute value changes its sign, so it is a critical point.
03

Evaluate at the endpoints

Evaluate the function at the endpoints of the interval. \(f(-3) = |6 - 4(-3)| = |6 + 12| = 18\) and \(f(3) = |6 - 4(3)| = |6 - 12| = 6\).
04

Evaluate at the critical point

Evaluate the function at the critical point \(x = 1.5\). \(f(1.5) = |6 - 4(1.5)| = |6 - 6| = 0\).
05

Compare values to find extrema

Now compare the obtained values: \(f(-3) = 18\), \(f(3) = 6\), and \(f(1.5) = 0\). The function achieves its absolute minimum value of 0 at \(x = 1.5\), and its absolute maximum value of 18 at \(x = -3\).

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Key Concepts

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

Critical Points
When dealing with functions, an important concept to grasp is critical points. These are specific values of the variable where the function's derivative is either zero or undefined.
For the function \(f(x) = |6 - 4x|\), critical points help us identify locations where the value of the function changes, often indicating maxima or minima.

Usually, with non-absolute functional forms, we find critical points by setting the derivative to zero. However, with absolute value functions, you need to be careful. The critical points can also be locations where the piecewise defined nature of the function changes.
In our case, \(x = 1.5\) is a critical point because it's where the expression \(6 - 4x\) inside the absolute value changes sign.
  • Critical points are vital in identifying the function's extrema within a given interval.
  • They help us pinpoint where the function's behavior shifts.
Absolute Value Function
The absolute value function \(|x|\) is one of the fundamental building blocks in mathematics, known for its "V"-shaped graph.
For any real number, the absolute value function turns negative numbers into their positive counterparts.
This feature makes it unique, and any function involving an absolute value has a natural piecewise definition.

In our exercise, the function \(f(x) = |6 - 4x|\) translates into two linear forms:
  • \(f(x) = 6 - 4x\) when \(x \leq 1.5\)
  • \(f(x) = -6 + 4x\) when \(x > 1.5\)
The point where the sign of the expression inside \(|...|\) changes is where the function shifts from one linear behavior to another.
Understanding these shifts are crucial because they delineate different regions where the function behaves differently, particularly during optimization problems.
Closed Interval
Closed intervals are an important concept when working with continuous functions. A closed interval, such as \([-3, 3]\), includes both of its endpoints.
This means that the function's values at these endpoints are just as important as values in between for finding absolute extrema.

In our example, evaluating the function \(f(x) = |6-4x|\) at the endpoints \(-3\) and \(3\) revealed that the extremal values could occur at these positions.
  • The function value at \(x = -3\) was found to be 18, indicating an absolute maximum.
  • At \(x = 3\), the value was 6, showing it as a candidate for extremum.
Being thorough in checking both critical points and endpoint values ensures that we correctly identify where a function's absolute maximum and minimum occur.

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Most popular questions from this chapter

The boundary of a field is a right triangle with a straight stream along its hypotenuse and with fences along its other two sides. Find the dimensions of the field with maximum area that can be enclosed using \(1000 \mathrm{ft}\) of fence.

Functions similar to $$ f(x)=\frac{1}{\sqrt{2 \pi}} e^{-x^{2} / 2} $$ arise in a wide variety of statistical problems. (a) Use the first derivative test to show that \(f\) has a relative maximum at \(x=0\), and confirm this by using a graphing utility to graph \(f\). (b) Sketch the graph of $$ f(x)=\frac{1}{\sqrt{2 \pi}} e^{-(x-\mu)^{2} / 2} $$ where \(\mu\) is a constant, and label the coordinates of the relative extrema.

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