Problem 16 - Calculus - AP Edition

Question

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

Step-by-Step Solution

Answer

The absolute maximum is 18 at \(x = -3\), and the absolute minimum is 0 at \(x = 1.5\).

Step 1: Understand the Absolute Function

The function given is \(f(x) = |6 - 4x|\). Absolute functions have a "V" shape and are split into two linear functions depending on whether the expression inside is positive or negative. In this case, \(f(x) = |6 - 4x|\) is split into two equations: \(f(x) = 6 - 4x\) when \(6 - 4x \geq 0\), or \(f(x) = 4x - 6\) when \(6 - 4x < 0\).

Step 2: Find the Critical Point

To find where the expression inside the absolute value changes from positive to negative, we solve the equation \(6 - 4x = 0\). Solving gives \(x = 1.5\). This is a critical point where the form of the function changes.

Step 3: Evaluate the Function at Critical Point and Endpoints

Evaluate \(f(x)\) at \(x = -3\), \(x = 1.5\), and \(x = 3\):- \(f(-3) = |6 - 4(-3)| = |6 + 12| = 18\)- \(f(1.5) = |6 - 4(1.5)| = |6 - 6| = 0\)- \(f(3) = |6 - 4(3)| = |6 - 12| = 6\)These calculations give us the function values at the critical point and endpoints of the interval.

Step 4: Determine Absolute Maximum and Minimum

From the values found:- Absolute maximum value is 18 at \(x = -3\).- Absolute minimum value is 0 at \(x = 1.5\).

Key Concepts

Critical Points

In the journey of finding absolute extrema, identifying critical points is key. Critical points are special values in the domain of a function where the behavior of the function changes. For absolute value functions, these are where the expression inside the absolute value becomes zero.

To find a critical point for the function \(f(x) = |6 - 4x|\), we set the expression inside the absolute value to zero: \(6 - 4x = 0\). Solving this equation, we find \(x = 1.5\) as the critical point.

Why is this important? At critical points, the graph of the function can change direction. For absolute value functions, they often represent the 'point' of the 'V' shape - where the slope flips. Understanding this notion is crucial for correctly determining absolute maxima and minima, especially within a closed interval where the function is continuous.

Absolute Value Function

Absolute value functions are fascinating due to their unique graph shape and properties. These functions transform all input values (the expressions within the absolute value) to their non-negative version.

For \(f(x) = |6 - 4x|\), the absolute value function can be viewed in two parts:

• \(f(x) = 6 - 4x\) when \(6 - 4x \geq 0\).
• \(f(x) = 4x - 6\) when \(6 - 4x < 0\).

This "V" shape is a defining feature, where the critical point at \(x = 1.5\) represents the vertex of the V. Analyzing each part separately helps in understanding how and where the function increases or decreases, which is important for pinpointing where the extreme values occur.
When dealing with absolute value functions, always be sure to consider the point where the absolute value expression equals zero, as well as the two linear pieces defined by it.

Closed Interval

When analyzing functions for maximum and minimum values, especially for absolute extrema, the closed interval is significant. A closed interval is expressed with square brackets \([-3, 3]\), meaning both endpoints are included. These endpoints are critical for determining absolute extrema.

For the function \(f(x) = |6 - 4x|\) over the closed interval \([-3, 3]\), we examine:

• Endpoints \(x = -3\) and \(x = 3\) for function values.
• Critical point \(x = 1.5\).

Evaluating \(f(x)\) at these points, we calculate:

• \(f(-3) = 18\)
• \(f(1.5) = 0\)
• \(f(3) = 6\)

Thus, on this closed interval, the function achieves an absolute maximum at \(x = -3\) and an absolute minimum at \(x = 1.5\).
The closed interval ensures these values are considered, as open intervals might exclude such evaluations at the endpoints, potentially missing where extrema occur. Always remember that the inclusion of endpoints is what differentiates closed intervals in this analysis.